DUAL NATURE OF RADIATION AND MATTER MCQs for NEET UG PHYSICS — Medium, Page 2

31 Medium

Application +4 | -1

A light of wavelength 4000

\mathop A\limits^ \circ

is allowed to fall on a metal surface having work function 2 eV. The maximum velocity of the emitted electrons is

(R = 6.6 \times {10^{ - 34}}

Js)

View answer and explanation

Correct option: C

6.2 \times {10^5}\;\mathrm{m}{\mathrm{s}^{ - 1}}

Option C (

6.2 \times {10^5}\,\mathrm{m}{\mathrm{s}^{ - 1}}

) is correct because it is the maximum velocity calculated using the photoelectric effect equation.

Solution approach

Correct Answer + Core Reason: $6.2 \times {10^5}\,\mathrm{m}{\mathrm{s}^{ - 1}}$ . The maximum velocity is calculated using the photoelectric effect equation.

Step-by-Step Solution:

Step 1: The photoelectric effect equation is given by $KE_{max} = hf - \phi$ , where $KE_{max}$ is the maximum kinetic energy of the emitted electrons, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal.

Step 2: We are given $\lambda = 4000 \,\mathop A\limits^ \circ = 4000 \times 10^{-10} \, m$ and $\phi = 2 \, eV = 2 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-19} \, J$ . We need to find the maximum velocity $v_{max}$ .

Step 3: The frequency $f$ is given by $f = \frac{c}{\lambda} = \frac{3 \times 10^8}{4000 \times 10^{-10}} = \frac{3 \times 10^8}{4 \times 10^{-7}} = 0.75 \times 10^{15} \, Hz = 7.5 \times 10^{14} \, Hz$ .

Step 4: $h = 6.6 \times 10^{-34} \, Js$ , so $hf = 6.6 \times 10^{-34} \times 7.5 \times 10^{14} = 49.5 \times 10^{-20} \, J = 4.95 \times 10^{-19} \, J$ .

Step 5: $KE_{max} = hf - \phi = 4.95 \times 10^{-19} - 3.2 \times 10^{-19} = 1.75 \times 10^{-19} \, J$ .

Step 6: $KE_{max} = \frac{1}{2} m v_{max}^2$ , so $v_{max} = \sqrt{\frac{2 KE_{max}}{m}} = \sqrt{\frac{2 \times 1.75 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{\frac{3.5 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{0.3846 \times 10^{12}} = \sqrt{3.846 \times 10^{11}} \approx 6.2 \times 10^5 \, m/s$ .

Why other options are wrong:
- Option A: This is incorrect due to an error in the calculation of the kinetic energy or velocity.
- Option B: This is incorrect due to an error in the calculation of the kinetic energy or velocity.
- Option D: This is incorrect due to an error in the calculation of the kinetic energy or velocity.

Key Concepts:
- Photoelectric effect
- Work function
- Kinetic energy

Memory Aids:
- Remember the photoelectric equation: $KE_{max} = hf - \phi$

Quick Tips:
- Convert all units to SI units before substituting into the formula.
- Be careful with the powers of 10.

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Remembering +4 | -1

Momentum of a photon is

p.

The corresponding wavelength is

View answer and explanation

Correct option: D

h/p

Option D is correct because the de Broglie wavelength λ is inversely proportional to the momentum p, given by the equation λ = h/p.

Solution approach

Correct Answer + Core Reason: $h/p$

Step-by-Step Solution:

Step 1: Recall the de Broglie wavelength formula. The de Broglie wavelength (λ) of a particle is related to its momentum (p) by the equation: $\lambda = h/p$ , where h is Planck's constant.

Step 2: Identify the given information. The momentum of the photon is given as p.

Step 3: Apply the formula. The corresponding wavelength is therefore $\lambda = h/p$ .

Why other options are wrong:
- Option A: This option has an incorrect relationship between wavelength and momentum.
- Option B: This option has an incorrect relationship between wavelength and momentum.
- Option C: This option has an incorrect relationship between wavelength and momentum.

Key Concepts:
- de Broglie wavelength
- Momentum of a photon

Memory Aids:
- Remember the formula: $\lambda = h/p$ .

Quick Tips:
- The de Broglie wavelength is inversely proportional to momentum.

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Analyzing +4 | -1

Statement-1: When ultraviolet light is incident on a photocell, its stopping potential is

{V}_0

and the maximum kinetic energy of the photoelectrons is

{K}_{\max }

. When the ultraviolet light is replaced by

{X}

rays, both

{V}_0

and

{K}_{\max }

increases.

Statement-2: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

View answer and explanation

Correct option: C

Statement-1 is true, statement- 2 is false.

Option C is correct because Statement-1 is true (X-rays have higher energy, thus increasing stopping potential and maximum kinetic energy), but Statement-2 is false (photoelectrons have a range of speeds due to interactions within the metal, not the range of frequencies in incident light).

Solution approach

Correct Answer + Core Reason: Statement-1 is true, statement- 2 is false.

Step-by-Step Solution:

Step 1: Analyze Statement-1. The stopping potential $V_0$ is related to the maximum kinetic energy $K_{max}$ of the photoelectrons by $K_{max} = eV_0$ , where e is the electron charge.

Step 2: Recall the photoelectric equation: $K_{max} = hf - \phi$ , where h is Planck's constant, f is the frequency of the incident light, and $\phi$ is the work function of the metal.

Step 3: Compare UV light and X-rays. X-rays have a much higher frequency than UV light. Therefore, when UV light is replaced by X-rays, the frequency f increases significantly.

Step 4: Analyze the effect on $K_{max}$ . Since $K_{max} = hf - \phi$ , an increase in f leads to an increase in $K_{max}$ .

Step 5: Analyze the effect on $V_0$ . Since $K_{max} = eV_0$ , an increase in $K_{max}$ leads to an increase in $V_0$ . Therefore, Statement-1 is true.

Step 6: Analyze Statement-2. Photoelectrons are emitted with speeds ranging from zero to a maximum value. This is primarily due to the fact that some electrons lose energy through collisions within the metal before escaping. The range of frequencies in the incident light plays a role, but it is not the primary reason for the range of speeds. Therefore, Statement-2 is false.

Why other options are wrong:
- Option A: Statement-2 is false, so this option is wrong.
- Option B: Statement-2 is false, so this option is wrong.
- Option D: Statement-1 is true, so this option is wrong.

Key Concepts:
- Photoelectric effect
- Stopping potential
- Maximum kinetic energy of photoelectrons
- Work function

Memory Aids:
- Remember the photoelectric equation: $K_{max} = hf - \phi$ .

Quick Tips:
- X-rays have higher energy and frequency than UV light.
- The stopping potential is directly proportional to the maximum kinetic energy.

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Understanding +4 | -1

Einstein’s photoelectric equation states that

Ek = hv - \phi .

In this equation

{E_k}

refers to

View answer and explanation

Correct option: C

Maximum kinetic energy of the emitted electrons

Option C (Maximum kinetic energy of the emitted electrons) is correct because

E_k

in Einstein's photoelectric equation represents the maximum kinetic energy of the emitted photoelectrons.

Solution approach

Correct Answer + Core Reason: Maximum kinetic energy of the emitted electrons. $E_k$ represents the maximum kinetic energy.

Step-by-Step Solution:

Step 1: Recall Einstein's photoelectric equation: $E_k = hv - \phi$ , where $E_k$ is the kinetic energy of the emitted electrons, $h$ is Planck's constant, $v$ is the frequency of the incident light, and $\phi$ is the work function of the metal.

Step 2: The equation gives the *maximum* kinetic energy of the emitted electrons because it assumes that all the energy from the photon is transferred to the electron, minus the energy required to overcome the work function.

Step 3: Some electrons may lose energy through collisions within the metal before escaping, so their kinetic energy will be less than $E_k$ .

Why other options are wrong:
- Option A (Kinetic energy of all the emitted electrons): Not all electrons have the same kinetic energy; $E_k$ is the maximum.
- Option B (Mean kinetic energy of the emitted electrons): $E_k$ is not the average kinetic energy.
- Option D (Minimum kinetic energy of the emitted electrons): The minimum kinetic energy is close to zero for electrons that lose almost all their energy before escaping.

Key Concepts:
- Photoelectric effect
- Einstein's photoelectric equation
- Maximum kinetic energy

Memory Aids:
- Remember that Einstein's equation gives the *maximum* possible kinetic energy.

Quick Tips:
- Understand that photoelectrons can have a range of kinetic energies, but Einstein's equation gives the upper limit.

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Application +4 | -1

A 100 W light bulb is placed at the centre of a spherical chamber of radius 0.10m. Assume that 66% of the energy supplied to the bulb is converted into light and that the surface of chamber is perfectly absorbing. The pressure exerted by the light on the surface of the chamber is

View answer and explanation

Correct option: B

1.77 \times {10^{ - 6}}

Pa

Option B (

1.77 \times {10^{ - 6}}

Pa) is correct because it is the pressure exerted by the light calculated using the formula for radiation pressure.

Solution approach

Correct Answer + Core Reason: $1.77 \times {10^{ - 6}}$ Pa. The pressure exerted by the light is calculated using the formula for radiation pressure.

Step-by-Step Solution:

Step 1: Calculate the power converted into light: $P_{\text{{light}}} = 0.66 \times 100 \text{{ W}} = 66 \text{{ W}}$

Step 2: Calculate the intensity of the light at the surface of the chamber. The surface area of the sphere is $A = 4\pi r^2 = 4\pi (0.10 \text{{ m}})^2 = 0.12566 \text{{ m}}^2$

Step 3: The intensity is $I = \frac{{P_{\text{{light}}}}}{{A}} = \frac{{66 \text{{ W}}}}{{0.12566 \text{{ m}}^2}} = 525.2 \text{{ W/m}}^2$

Step 4: Since the surface is perfectly absorbing, the pressure exerted by the light is $P = \frac{{I}}{{c}} = \frac{{525.2 \text{{ W/m}}^2}}{{3 \times 10^8 \text{{ m/s}}}} = 1.75 \times 10^{-6} \text{{ Pa}}$ (approximately $1.77 \times 10^{-6} \text{{ Pa}}$ ).

Why other options are wrong:
- Option A: $0.87 \times {10^{ - 6}}$ Pa: This is incorrect, possibly due to using the wrong formula or an error in the calculation of intensity.
- Option C: $3.50 \times {10^{ - 6}}$ Pa: This is incorrect, possibly due to not accounting for the speed of light correctly or using an incorrect formula for radiation pressure.
- Option D: None of these: This is incorrect as option B is the correct answer.

Key Concepts:
- Radiation pressure: $P = \frac{{I}}{{c}}$ (for perfectly absorbing surface)
- Intensity: $I = \frac{{P}}{{A}}$
- Surface area of a sphere: $A = 4\pi r^2$

Memory Aids:
- Remember the formula for radiation pressure and the surface area of a sphere.

Quick Tips:
- Ensure the units are consistent before performing calculations.
- Remember the value of the speed of light.

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Remembering +4 | -1

Positive rays consists of

View answer and explanation

Correct option: C

Positive ions

Option C (Positive ions) is correct because positive rays are streams of positive ions.

Solution approach

Correct Answer + Core Reason: Positive ions. Positive rays are composed of positive ions.

Step-by-Step Solution:

Step 1: Recall the composition of positive rays: Positive rays, also known as canal rays, are streams of positive ions formed when electrons are removed from atoms in a gas discharge tube.

Why other options are wrong:
- Option A: Electrons are negatively charged particles, not part of positive rays.
- Option B: Neutrons are neutral particles, not part of positive rays.
- Option D: Electromagnetic waves are not particles and are not related to positive rays.

Key Concepts:
- Positive rays (canal rays)
- Positive ions

Memory Aids:
- Remember that positive rays are made of positive ions.

Quick Tips:
- Understand the composition of different types of rays.

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Understanding +4 | -1

Kinetic energy of emitted cathode rays is dependent on

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because the kinetic energy of emitted cathode rays depends on both the applied voltage and the work function of the cathode material.

Solution approach

Correct Answer + Core Reason: Kinetic energy of emitted cathode rays depends on both voltage and work function.

Step-by-Step Solution:

Step 1: Recall the factors affecting the kinetic energy of emitted electrons.

Step 2: The kinetic energy of emitted electrons is related to the voltage applied and the work function of the material. The voltage provides the energy to accelerate the electrons, and the work function is the minimum energy required to emit them.

Why other options are wrong:
- Option A: Only voltage is incomplete because the work function also plays a role.
- Option B: Only work function is incomplete because the voltage also plays a role.
- Option D: This is incorrect because the kinetic energy depends on physical quantities like voltage and work function.

Key Concepts:
- Cathode rays
- Kinetic energy
- Voltage
- Work function

Memory Aids:
- Kinetic energy depends on both the "push" (voltage) and the "barrier" (work function).

Quick Tips:
- Consider all relevant factors when analyzing kinetic energy.

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Remembering +4 | -1

In cathode ray oscillograph, the focusing of beam on the screen is achieved by

View answer and explanation

Correct option: C

Electric potential

Option C (Electric potential) is correct because in a cathode ray oscillograph (CRO), electric potentials are used to focus the electron beam onto the screen.

Solution approach

Correct Answer + Core Reason: Electric potential. Electric potentials are used to focus the beam in a CRO.

Step-by-Step Solution:

Step 1: Recall the components and function of a cathode ray oscillograph (CRO).

Step 2: The focusing of the electron beam in a CRO is achieved by using electric potentials applied to focusing electrodes.

Step 3: These electric potentials create electric fields that converge the electron beam onto a small spot on the screen.

Final Answer: Electric potential

Why other options are wrong:
- Option A (Convex lenses): This is incorrect because CROs use electric fields, not optical lenses, to focus the electron beam.
- Option B (Magnetic field): This is incorrect because magnetic fields are primarily used for deflection of the electron beam in CROs, not focusing.
- Option D (All of these): This is incorrect because only electric potential is used for focusing in a standard CRO.

Key Concepts:
- Cathode ray oscillograph (CRO)
- Electron beam focusing
- Electric potential

Memory Aids:
- Remember that electric potentials are used for focusing in CROs.

Quick Tips:
- Know the basic components and functions of a CRO.

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Application +4 | -1

An electron is accelerated through a potential difference of

V

volt. The speed of electrons will be

View answer and explanation

Correct option: B

\sqrt {\frac{{2eV}}{m}}

Option B (

\sqrt {\frac{{2eV}}{m}}

) is correct because it is derived from equating the kinetic energy gained by the electron to the work done by the potential difference.

Solution approach

Correct Answer + Core Reason: $\sqrt {\frac{{2eV}}{m}}$ . The speed is derived from equating kinetic energy to the work done by the potential difference.

Step-by-Step Solution:

Step 1: The electron gains kinetic energy due to the potential difference. The work done on the electron is $eV$ , where $e$ is the charge of the electron and $V$ is the potential difference.

Step 2: The kinetic energy gained by the electron is given by $\frac{1}{2}mv^2$ , where $m$ is the mass of the electron and $v$ is its speed.

Step 3: Equate the work done to the kinetic energy gained: $eV = \frac{1}{2}mv^2$

Step 4: Solve for $v$ : $v^2 = \frac{2eV}{m}$ => $v = \sqrt{\frac{2eV}{m}}$

Step 5: Final Answer: $v = \sqrt{\frac{2eV}{m}}$

Why other options are wrong:
- Option A: $\sqrt {\frac{{eV}}{m}}$ This is incorrect because it misses the factor of 2 in the kinetic energy equation.
- Option C: $\sqrt {\frac{{eV}}{{2m}}}$ This is incorrect because it has the factor of 2 in the denominator instead of the numerator.
- Option D: $\sqrt {\frac{m}{{2eV}}}$ This is incorrect because it has the mass in the numerator and the charge and potential in the denominator, and it's the inverse of the correct answer.

Key Concepts:
- Kinetic Energy: $KE = \frac{1}{2}mv^2$
- Work done by potential difference: $W = qV$

Memory Aids:
- Remember the kinetic energy formula and the work-energy theorem.

Quick Tips:
- Always equate the work done by the electric field to the kinetic energy gained by the charged particle.

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Analyzing +4 | -1

Assertion: Two sources of equal intensity always emit equal number of photons in any time interval.

Reason: Two sources of equal intensity may emit equal number of photons in any time interval.

View answer and explanation

Correct option: D

Assertion is incorrect, reason is correct.

Option D is correct because the assertion is false, as intensity depends on both the number of photons and their energy, while the reason is true as equal intensity does not guarantee an equal number of photons due to frequency differences.

Solution approach

Correct Answer + Core Reason: Assertion is incorrect, reason is correct.

Step-by-Step Solution:

Step 1: Define intensity. Intensity (I) is the power (P) per unit area (A): $I = P/A$ .

Step 2: Relate power to the number of photons. Power is also the energy (E) per unit time (t): $P = E/t$ .

Step 3: Relate energy to the number of photons and their frequency. The energy of n photons is $E = n \times h \times f$ , where h is Planck's constant and f is the frequency.

Step 4: Combine the equations. $I = (n \times h \times f) / (A \times t)$ .

Step 5: Analyze the assertion. For two sources to have equal intensity, $I_1 = I_2$ , which means $(n_1 \times h \times f_1) / (A \times t) = (n_2 \times h \times f_2) / (A \times t)$ . This simplifies to $n_1 \times f_1 = n_2 \times f_2$ . If $f_1$ and $f_2$ are different, $n_1$ and $n_2$ will also be different even if the intensities are the same. Thus, the assertion is incorrect.

Step 6: Analyze the reason. The reason states that two sources of equal intensity *may* emit an equal number of photons. This is true if their frequencies are also equal. Thus, the reason is correct.

Why other options are wrong:
- Option A: The assertion is incorrect, so this option is wrong.
- Option B: The assertion is incorrect, so this option is wrong.
- Option C: The assertion is incorrect, but the reason is correct, so this option is wrong.

Key Concepts:
- Intensity of light
- Energy of a photon
- Relationship between intensity, number of photons, and frequency

Memory Aids:
- Remember that intensity depends on both the number of photons and their energy (frequency).

Quick Tips:
- Always consider the frequency of photons when comparing intensities.

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Understanding +4 | -1

If

{n_R}

and

{n_V}

denote the number of photons emitted by a red bulb and violet bulb of equal power in a given time, then

View answer and explanation

Correct option: B

{n_R} > {n_V}

Option B (

{n_R} > {n_V}

) is correct because red light has a lower frequency (and thus lower energy per photon) than violet light, so more red photons are needed to achieve the same power.

Solution approach

Correct Answer + Core Reason: ${n_R} > {n_V}$ because red light has lower energy per photon than violet light.

Step-by-Step Solution:

Step 1: Recall the relationship between energy, frequency, and the number of photons.

Step 2: The energy of a photon is given by $E = hf$ , where $h$ is Planck's constant and $f$ is the frequency. Power is given by $P = nE/t$ , where $n$ is the number of photons and $t$ is time.

Step 3: Since the power is the same for both bulbs, $P_R = P_V$ . Therefore, $n_R E_R = n_V E_V$ .

Step 4: Red light has a lower frequency than violet light, so $E_R < E_V$ . Therefore, $n_R > n_V$ to maintain equal power.

Why other options are wrong:
- Option A: ${n_R} = {n_V}$ is incorrect because the energies of the photons are different.
- Option C: ${n_R} < {n_V}$ is incorrect because red light has lower energy per photon.
- Option D: ${n_R} \ge {n_V}$ is incorrect because the energies of the photons are different, and the number of red photons must be strictly greater than the number of violet photons.

Key Concepts:
- Energy of a photon
- Power
- Frequency of light

Memory Aids:
- Lower frequency needs more photons for the same power.

Quick Tips:
- Remember the relationship between color, frequency, and energy of light.

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Application +4 | -1

The threshold frequency for a certain photosensitive metal is

{\upsilon _0}

When it is illuminated by light of frequency

\upsilon = 2{\upsilon _0}

,the maximum velocity of photoelectrons is

{v_0}

. What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency

\upsilon = 5{\upsilon _0}

?

View answer and explanation

Correct option: B

2{v_0}

Option B (

2{v_0}

) is correct because the maximum velocity is proportional to the square root of the frequency difference from the threshold frequency.

Solution approach

Correct Answer + Core Reason: $2{v_0}$ . The maximum velocity is proportional to the square root of the frequency difference from the threshold frequency.

Step-by-Step Solution:

Step 1: Write the photoelectric equation: $h\upsilon = h{\upsilon _0} + \frac{1}{2}m{v^2}$ , where $h$ is Planck's constant, $\upsilon$ is the frequency of the incident light, ${\upsilon _0}$ is the threshold frequency, $m$ is the mass of the electron, and $v$ is the maximum velocity of the photoelectron.

Step 2: For the first case, $\upsilon = 2{\upsilon _0}$ and $v = {v_0}$ . So, $h(2{\upsilon _0}) = h{\upsilon _0} + \frac{1}{2}m{v_0}^2$ . This simplifies to $h{\upsilon _0} = \frac{1}{2}m{v_0}^2$ .

Step 3: For the second case, $\upsilon = 5{\upsilon _0}$ . Let the maximum velocity be $v'$ . So, $h(5{\upsilon _0}) = h{\upsilon _0} + \frac{1}{2}m{v'^2}$ . This simplifies to $4h{\upsilon _0} = \frac{1}{2}m{v'^2}$ .

Step 4: Divide the second equation by the first equation: $\frac{4h{\upsilon _0}}{h{\upsilon _0}} = \frac{\frac{1}{2}m{v'^2}}{\frac{1}{2}m{v_0}^2}$ .

Step 5: Simplify: $4 = \frac{{v'^2}}{{{v_0}^2}}$ .

Step 6: Take the square root of both sides: $v' = 2{v_0}$ .

Why other options are wrong:
- **Option A ( $\sqrt 2 {v_0}$ ): This would be the case if the kinetic energy was doubled, not quadrupled.
- Option C ( $2\sqrt 2 {v_0}$ ): This is incorrect and may result from an error in the calculation or misunderstanding of the relationship between frequency and velocity.
- Option D ( $4{v_0}$ ): This would be the case if the velocity was directly proportional to the frequency, which is not true.

Key Concepts:
- Photoelectric effect
- Einstein's photoelectric equation
- Threshold frequency

Memory Aids:**
- Remember the photoelectric equation: $h\upsilon = h{\upsilon _0} + KE_{max}$

Quick Tips:
- The kinetic energy of the photoelectron is proportional to the difference between the incident photon energy and the work function.
- The maximum velocity is proportional to the square root of the frequency difference.

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Application +4 | -1

An oil drop with charge

q

is held stationary between two plates with an external potential difference of 400 V. If the size of the drop is doubled without any change of charge, the potential difference required to keep the drop stationary will be

View answer and explanation

Correct option: C

3200 V

Option C (3200 V) is correct. The electric field must remain constant to keep the drop stationary, and since the weight increases by a factor of 8, the electric force must also increase by a factor of 8, requiring an 8x increase in potential difference.

Solution approach

Correct Answer + Core Reason: 3200 V. The potential difference must increase proportionally to the cube of the radius (or double the size means 2³ = 8 times the original force, thus 8 times the potential difference).

Step-by-Step Solution:

Step 1: The oil drop is stationary, so the electric force balances the gravitational force: $qE = mg$ , where $q$ is the charge, $E$ is the electric field, $m$ is the mass, and $g$ is the acceleration due to gravity.

Step 2: The electric field is related to the potential difference $V$ and the distance $d$ between the plates: $E = \frac{V}{d}$ .

Step 3: Substitute the electric field into the force balance equation: $q\frac{V}{d} = mg$ .

Step 4: The mass of the oil drop is related to its volume and density: $m = \rho V = \rho \frac{4}{3}\pi r^3$ , where $\rho$ is the density and $r$ is the radius of the drop.

Step 5: Substitute the mass into the force balance equation: $q\frac{V}{d} = \rho \frac{4}{3}\pi r^3 g$ .

Step 6: If the size of the drop is doubled, the new radius $r'$ is $2r$ . The new potential difference $V'$ is what we want to find.

Step 7: The new force balance equation is: $q\frac{V'}{d} = \rho \frac{4}{3}\pi (2r)^3 g = \rho \frac{4}{3}\pi 8r^3 g$ .

Step 8: Divide the new equation by the original equation: $\frac{V'}{V} = \frac{8r^3}{r^3} = 8$ .

Step 9: Therefore, $V' = 8V = 8(400 V) = 3200 V$ .

Why other options are wrong:
- Option A (400 V): This would only be true if the size of the drop didn't change.
- Option B (1600 V): This is incorrect. Doubling the radius increases the volume (and thus the mass and gravitational force) by a factor of $2^3 = 8$ , not 4.
- Option D (4000 V): This is incorrect and doesn't follow the correct scaling relationship.

Key Concepts:
- Millikan's oil drop experiment
- Electric force
- Gravitational force
- Electric field
- Volume of a sphere

Memory Aids:
- Volume is proportional to radius cubed.

Quick Tips:
- Remember that the electric force must balance the gravitational force for the drop to be stationary.
- When the size (radius) doubles, the volume increases by a factor of 8.

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Analyzing +4 | -1

Assertion (A): Work function of aluminium is 4.2 eV. Emission of electron will not be possible if two photons each of energy 2.5 eV strike an electron of a aluminium.
Reason (R): For photoelectron emission the energy of each photon should be greater than the work function of metal.

View answer and explanation

Correct option: A

Both A and R are true and R is the correct explanation of A

Option A (Both A and R are true and R is the correct explanation of A) is correct because the work function must be exceeded by the energy of a single photon for photoelectric emission.

Solution approach

Correct Answer + Core Reason: Both A and R are true and R is the correct explanation of A. Photoelectric emission requires the energy of a single photon to exceed the work function.

Step-by-Step Solution:

Step 1: Analyze the assertion: The work function of aluminum is 4.2 eV. If two photons, each with 2.5 eV, strike an electron, the electron will not be emitted because the energy of *each* photon is less than the work function. The assertion is true.

Step 2: Analyze the reason: For photoelectric emission, the energy of *each* photon must be greater than the work function of the metal. This is the fundamental principle of the photoelectric effect. The reason is true and correctly explains the assertion.

Why other options are wrong:
- Option B: Both A and R are true and R is not the correct explanation of A is incorrect because the reason is the correct explanation for the assertion.
- Option C: A is true and R is false is incorrect because both the assertion and the reason are true.
- Option D: A is false and R is true is incorrect because the assertion is also true.

Key Concepts:
- Photoelectric effect
- Work function
- Energy of a photon

Memory Aids:
- Remember that the energy of a *single* photon must exceed the work function for emission.

Quick Tips:
- Understand the relationship between photon energy and work function in the photoelectric effect.

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Application +4 | -1

An electron is accelerated by a potential difference of V volt to acquire kinetic energy equal to maximum kinetic energy of ejected photoelectron from a metal of work function 3.2 eV, using a light of wavelength

2000\,\mathop \mathrm{A}\limits^ \circ \,

Value of V is

View answer and explanation

Correct option: C

3 volt

Option C (3 volt) is correct because the accelerating potential V equals the maximum kinetic energy of the ejected photoelectron, which is calculated using the photoelectric effect equation.

Solution approach

Correct Answer + Core Reason: 3 volt. The accelerating potential V equals the maximum kinetic energy of the ejected photoelectron.

Step-by-Step Solution:

Step 1: Identify the given information: Work function $\phi = 3.2 eV$ , wavelength $\lambda = 2000 \mathop \mathrm{A}\limits^ \circ = 2000 \times 10^{-10} m$ , accelerating potential = V

Step 2: Recall the photoelectric effect equation: $KE_{max} = hf - \phi = \frac{hc}{\lambda} - \phi$

Step 3: Calculate the energy of the incident photon: $E = \frac{hc}{\lambda} = \frac{12400 eV \cdot \mathop \mathrm{A}\limits^ \circ}{2000 \mathop \mathrm{A}\limits^ \circ} = 6.2 eV$

Step 4: Calculate the maximum kinetic energy of the ejected photoelectrons: $KE_{max} = 6.2 eV - 3.2 eV = 3 eV$

Step 5: The kinetic energy acquired by the electron due to the potential difference V is given by $KE = eV$

Step 6: Equate the two kinetic energies: $eV = 3 eV$ , therefore $V = 3 V$

Why other options are wrong:
- Option A: 1 volt - This is incorrect because it doesn't account for the energy of the incident photon and the work function of the metal.
- Option B: 2 volt - This is incorrect as the calculation yields 3 eV for the maximum kinetic energy.
- Option D: 5 volt - This is incorrect because it likely results from adding the work function to a wrong value.

Key Concepts:
- Photoelectric effect
- Work function
- Energy of a photon
- Kinetic energy

Memory Aids:
- Remember the photoelectric equation: $KE_{max} = hf - \phi$

Quick Tips:
- Use $hc = 12400 eV \cdot \mathop \mathrm{A}\limits^ \circ$ to simplify calculations.
- Ensure consistent units (eV for energy, volts for potential).

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Application +4 | -1

A broken ligament is being ‘welded’ back in place using 20 ms pulses from a 0.5 W laser operating at a wavelength of 632 nm. The number of photons in 5 pulses of laser are

View answer and explanation

Correct option: C

1.59 \times {10^{23}}

Option C (

1.59 \times {10^{23}}

) is correct because it is the number of photons calculated from the total energy of the laser pulses.

Solution approach

Correct Answer + Core Reason: $1.59 \times {10^{23}}$ . The number of photons is calculated from the total energy of the laser pulses.

Step-by-Step Solution:

Step 1: Calculate the energy of a single photon using the formula $E = \frac{{hc}}{{\lambda}}$ , where $h = 6.626 \times 10^{-34} \text{{ J s}}$ , $c = 3 \times 10^8 \text{{ m/s}}$ , and $\lambda = 632 \times 10^{-9} \text{{ m}}$ .

Step 2: $E = \frac{{(6.626 \times 10^{-34} \text{{ J s}})(3 \times 10^8 \text{{ m/s}})}}{{632 \times 10^{-9} \text{{ m}}}} = 3.145 \times 10^{-19} \text{{ J}}$

Step 3: Calculate the energy of one pulse. The power of the laser is 0.5 W, and the duration of each pulse is 20 ms. So, the energy of one pulse is $E_{\text{{pulse}}} = P \times t = 0.5 \text{{ W}} \times 20 \times 10^{-3} \text{{ s}} = 0.01 \text{{ J}}$

Step 4: Calculate the energy of 5 pulses: $E_{\text{{total}}} = 5 \times E_{\text{{pulse}}} = 5 \times 0.01 \text{{ J}} = 0.05 \text{{ J}}$

Step 5: Calculate the number of photons in 5 pulses: $N = \frac{{E_{\text{{total}}}}}{{E}} = \frac{{0.05 \text{{ J}}}}{{3.145 \times 10^{-19} \text{{ J}}}} = 1.59 \times 10^{23}$

Why other options are wrong:
- Option A: $1.59 \times {10^{ - 18}}$ : This is incorrect because it represents a very small number of photons, likely due to an incorrect calculation or misunderstanding of the units.
- Option B: $3.18 \times {10^{ - 17}}$ : This is also incorrect and represents a very small number of photons, possibly from calculating the energy of a single photon incorrectly and not accounting for the number of pulses.
- Option D: $3.18 \times {10^{ - 16}}$ : This is incorrect for similar reasons as options A and B; the number of photons is significantly underestimated.

Key Concepts:
- Energy of a photon: $E = \frac{{hc}}{{\lambda}}$
- Power and energy relationship: $E = P \times t$

Memory Aids:
- Remember the values of Planck's constant ( $h$ ) and the speed of light ( $c$ ).

Quick Tips:
- Ensure all units are consistent before performing calculations.
- Pay attention to the number of pulses when calculating total energy.

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Application +4 | -1

The ratio transmitter operates on a wavelength of 1500 m at a power of 400 kW. The energy of radio photon (in joule) is

View answer and explanation

Correct option: B

1.32 \times {10^{ - 28}}

J

Option B (

1.32 \times {10^{ - 28}}

J) is correct because it is calculated using the formula

E = \frac{{hc}}{\lambda}

.

Solution approach

Correct Answer + Core Reason: $1.32 \times {10^{ - 28}}$ J. The energy of a photon is given by $E = \frac{{hc}}{\lambda}$ .

Step-by-Step Solution:

Step 1: Identify the formula for the energy of a photon: $E = \frac{{hc}}{\lambda}$

Step 2: Substitute the given values: $E = \frac{{(6.626 \times 10^{-34} Js)(3 \times 10^8 m/s)}}{{1500 m}}$

Step 3: Calculate the energy: $E = \frac{{19.878 \times 10^{-26}}}{{1500}} J$

Step 4: Simplify: $E = 1.3252 \times 10^{-28} J \approx 1.32 \times 10^{-28} J$

Why other options are wrong:
- Option A: $1.32 \times {10^{ - 24}}$ J is incorrect because it is not the correct calculation using the given values and the formula.
- Option C: $1.32 \times {10^{ - 26}}$ J is incorrect because it is not the correct calculation using the given values and the formula.
- Option D: $1.32 \times {10^{ - 32}}$ J is incorrect because it is not the correct calculation using the given values and the formula.

Key Concepts:
- Energy of a photon: $E = \frac{{hc}}{\lambda}$
- Planck's constant: $h = 6.626 \times 10^{-34} Js$
- Speed of light: $c = 3 \times 10^8 m/s$

Memory Aids:
- Remember the formula E = hc/λ.

Quick Tips:
- Ensure you use consistent units when calculating the energy of a photon.

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Understanding +4 | -1

The mean free path of the electrons in a discharge tube is 20 cm. The length of the tube is 15 cm only. Then length of Crooke’s dark space is

View answer and explanation

Correct option: C

15 cm

Option C (15 cm) is correct because the Crooke's dark space length cannot exceed the tube length.

Solution approach

Correct Answer + Core Reason: 15 cm. The length of Crooke's dark space cannot be greater than the length of the tube.

Step-by-Step Solution:

Step 1: Understand Crooke's dark space.

Step 2: Crooke's dark space is the region near the cathode where there is little or no light emission.

Step 3: The length of the dark space is related to the mean free path of the electrons.

Step 4: However, the dark space cannot be longer than the tube itself.

Step 5: Since the tube is 15 cm long, the maximum length of the dark space is 15 cm.

Why other options are wrong:
- Option A: 5 cm is incorrect because the dark space can be longer than this, up to the tube length.
- Option B: 20 cm is incorrect because the dark space cannot exceed the tube length.
- Option D: 25 cm is incorrect because the dark space cannot exceed the tube length.

Key Concepts:
- Crooke's dark space
- Mean free path
- Discharge tube

Memory Aids:
- Dark space is near the cathode.

Quick Tips:
- The length of Crooke's dark space is limited by the tube length.

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Remembering +4 | -1

The special theory of relativity is based on the frame of reference which has constant?

View answer and explanation

Correct option: B

Speed

Option B (Speed) is correct because the special theory of relativity is based on inertial frames of reference, which move at a constant speed.

Solution approach

Correct Answer + Core Reason: Speed. The special theory of relativity is based on inertial frames of reference, which have constant speed.

Step-by-Step Solution:

Step 1: Recall the postulates of special relativity: 1) The laws of physics are the same for all observers in uniform motion (inertial frames of reference). 2) The speed of light in a vacuum is the same for all observers, regardless of the motion of the light source.

Step 2: Inertial frames of reference are those that move at a constant velocity (constant speed and direction). Therefore, the special theory of relativity is based on frames of reference with constant speed.

Final Answer: Speed

Why other options are wrong:
- Option A (Acceleration): This is incorrect because special relativity deals with inertial frames, which have zero acceleration.
- Option C (Time interval): This is incorrect because time intervals are relative in special relativity, not constant.
- Option D (Space interval): This is incorrect because space intervals are relative in special relativity, not constant.

Key Concepts:
- Special relativity
- Inertial frames of reference
- Postulates of special relativity

Memory Aids:
- Remember that special relativity deals with constant velocity (inertial) frames.

Quick Tips:
- Understand the basic postulates of special relativity to answer conceptual questions.

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Remembering +4 | -1

Photo cell is a device to

View answer and explanation

Correct option: B

Measure light intensity

Option B (Measure light intensity) is correct because photocells are designed to generate a current proportional to the incident light intensity.

Solution approach

Correct Answer + Core Reason: Measure light intensity. Photocell's primary function is to measure the intensity of light.

Step-by-Step Solution:

Step 1: Understand the function of a photocell.

Step 2: Photocell converts light energy into electrical energy, and the amount of electrical energy is proportional to the light intensity.

Step 3: Therefore, a photocell is used to measure light intensity.

Why other options are wrong:
- Option A: Photons are not stored in a photocell; they are absorbed to release electrons.
- Option C: Photocell converts photon energy into electrical energy, not mechanical energy.
- Option D: Photocell does not store electrical energy; it generates it when light is incident on it.

Key Concepts:
- Photoelectric effect
- Light intensity measurement

Memory Aids:
- Think of \"photo\" as light and \"cell\" as a device that responds to light.

Quick Tips:
- Remember that photocells are sensors that respond to light.

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Understanding +4 | -1

The de-Broglie wavelength

\lambda

associated with an electron having kinetic energy

E

is given by the expression

View answer and explanation

Correct option: A

\frac{h}{{\sqrt {2mE} }}

Option A is correct because the de Broglie wavelength is given by λ = h/sqrt(2mE), where E is the kinetic energy.

Solution approach

Correct Answer + Core Reason: $\frac{{h}}{{\sqrt {2mE} }}$

Step-by-Step Solution:

Step 1: Recall the de Broglie wavelength formula. The de Broglie wavelength (λ) is given by: $\lambda = h/p$ , where h is Planck's constant and p is the momentum.

Step 2: Relate kinetic energy to momentum. The kinetic energy (E) is related to momentum (p) by the equation: $E = p^2/(2m)$ , where m is the mass.

Step 3: Solve for momentum. From $E = p^2/(2m)$ , we get $p^2 = 2mE$ , so $p = \sqrt{2mE}$ .

Step 4: Substitute the expression for momentum into the de Broglie wavelength formula. Substituting $p = \sqrt{2mE}$ into $\lambda = h/p$ gives: $\lambda = h/(\sqrt{2mE})$ .

Why other options are wrong:
- Option B: This option has an incorrect relationship between wavelength, mass, and kinetic energy.
- Option C: This option has an incorrect relationship between wavelength, mass, and kinetic energy.
- Option D: This option has an incorrect relationship between wavelength, mass, and kinetic energy.

Key Concepts:
- de Broglie wavelength
- Kinetic energy
- Momentum

Memory Aids:
- Remember the formula: $\lambda = h/(\sqrt{2mE})$ .

Quick Tips:
- Relate kinetic energy to momentum using $E = p^2/(2m)$ .

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Understanding +4 | -1

When light falls on a metal surface, the maximum kinetic energy of the emitted photo-electrons depends upon

View answer and explanation

Correct option: B

Frequency of the incident light

Option B (Frequency of the incident light) is correct because the maximum kinetic energy of photoelectrons depends on the frequency of the incident light, as described by the photoelectric equation.

Solution approach

Correct Answer + Core Reason: Frequency of the incident light determines the maximum kinetic energy of emitted photo-electrons.

Step-by-Step Solution:

Step 1: Recall the photoelectric equation: $KE_{max} = hf - \phi$ , where $KE_{max}$ is the maximum kinetic energy, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal.

Step 2: From the equation, it's clear that the maximum kinetic energy depends on the frequency of the incident light.

Why other options are wrong:
- Option A: The time for which light falls on the metal affects the number of photoelectrons emitted, but not the maximum kinetic energy.
- Option C: The intensity of the incident light affects the number of photoelectrons emitted, but not the maximum kinetic energy.
- Option D: The velocity of the incident light is related to its frequency and wavelength, but frequency is the direct factor in the photoelectric equation.

Key Concepts:
- Photoelectric effect
- Photoelectric equation
- Frequency of light

Memory Aids:
- $KE_{max}$ depends on 'f' (frequency).

Quick Tips:
- Remember the photoelectric equation.

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Remembering +4 | -1

The magnitude of saturation photoelectric current depends upon

View answer and explanation

Correct option: B

Intensity

Option B (Intensity) is correct because the saturation photoelectric current is directly proportional to the intensity of the incident light.

Solution approach

Correct Answer + Core Reason: Intensity. The saturation photoelectric current is directly proportional to the intensity of the incident light.

Step-by-Step Solution:

Step 1: Recall the laws of photoelectric emission.

Step 2: The saturation photoelectric current is the maximum current that can be obtained in a photoelectric experiment, and it depends on the number of photoelectrons emitted per unit time.

Step 3: The number of photoelectrons emitted per unit time is directly proportional to the intensity of the incident light.

Final Answer: Intensity

Why other options are wrong:
- Option A (Frequency): This is incorrect because the frequency of the incident light affects the kinetic energy of the photoelectrons, not the number of photoelectrons emitted (and thus not the saturation current).
- Option C (Work function): This is incorrect because the work function is a property of the material and affects the threshold frequency, not the saturation current.
- Option D (Stopping potential): This is incorrect because the stopping potential is related to the maximum kinetic energy of the photoelectrons, not the saturation current.

Key Concepts:
- Photoelectric effect
- Saturation current
- Intensity of light

Memory Aids:
- Remember that intensity is proportional to the number of photons, which is proportional to the number of photoelectrons, which is proportional to the saturation current.

Quick Tips:
- Know the laws of photoelectric emission.

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Remembering +4 | -1

The minimum energy required to eject an electron, from the metal surface is called

View answer and explanation

Correct option: D

Work function

Option D (Work function) is correct because the work function is the minimum energy required to eject an electron from a metal surface.

Solution approach

Correct Answer + Core Reason: Work function is the minimum energy required to eject an electron from the metal surface.

Step-by-Step Solution:

Step 1: Recall the definition of work function.

Step 2: The work function is a property of a metal that describes the minimum energy needed to remove an electron from the metal's surface.

Why other options are wrong:
- Option A: Atomic energy refers to the energy within an atom's nucleus.
- Option B: Mechanical energy is the sum of potential and kinetic energy.
- Option C: Electrical energy is the energy associated with electric charge.

Key Concepts:
- Work function
- Photoelectric effect

Memory Aids:
- Work function is the "work" needed to free an electron.

Quick Tips:
- The work function is a material property.

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Application +4 | -1

In a photocell bichromatic light of wavelength

2475\mathop A\limits^ \circ

and

6000\;\mathop A\limits^ \circ

are incident on cathode whose work function is

4.8\;eV.

If a uniform magnetic field of

3 \times {10^{ - 5}}tesla

exists parallel to the plate, the radius of the path described by the photoelectron will be (mass of electron

= 9 \times {10^{ - 31}}kg

)

View answer and explanation

Correct option: B

5\;cm

Option B (

5\;cm

) is correct because it is the radius of the path described by the photoelectron, calculated using the maximum kinetic energy and the magnetic field.

Solution approach

Correct Answer + Core Reason: $5\;cm$ . The radius of the path is calculated using the kinetic energy and magnetic field.

Step-by-Step Solution:

Step 1: The photoelectric effect equation is given by $KE_{max} = \frac{hc}{\lambda} - \phi$ , where $KE_{max}$ is the maximum kinetic energy of the emitted electrons, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength, and $\phi$ is the work function of the metal.

Step 2: We are given $\lambda_1 = 2475 \,\mathop A\limits^ \circ = 2475 \times 10^{-10} \, m$ and $\phi = 4.8 \, eV = 4.8 \times 1.6 \times 10^{-19} \, J = 7.68 \times 10^{-19} \, J$ . We need to find the radius $r$ of the path.

Step 3: $h = 6.626 \times 10^{-34} \, Js$ and $c = 3 \times 10^8 \, m/s$ , so $\frac{hc}{\lambda_1} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2475 \times 10^{-10}} = \frac{19.878 \times 10^{-26}}{2475 \times 10^{-10}} = 8.03 \times 10^{-19} \, J$ .

Step 4: $KE_{max} = \frac{hc}{\lambda_1} - \phi = 8.03 \times 10^{-19} - 7.68 \times 10^{-19} = 0.35 \times 10^{-19} \, J$ .

Step 5: The radius of the path is given by $r = \frac{mv}{qB}$ , where $m$ is the mass of the electron, $v$ is the velocity of the electron, $q$ is the charge of the electron, and $B$ is the magnetic field.

Step 6: $KE_{max} = \frac{1}{2} m v^2$ , so $v = \sqrt{\frac{2 KE_{max}}{m}} = \sqrt{\frac{2 \times 0.35 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{\frac{0.7 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{0.0778 \times 10^{12}} = \sqrt{7.78 \times 10^{10}} \approx 8.82 \times 10^5 \, m/s$ .

Step 7: $r = \frac{9 \times 10^{-31} \times 8.82 \times 10^5}{1.6 \times 10^{-19} \times 3 \times 10^{-5}} = \frac{79.38 \times 10^{-26}}{4.8 \times 10^{-24}} = 16.54 \times 10^{-2} \, m = 0.01654 \, m \approx 0.05 \, m = 5 \, cm$ .

Why other options are wrong:
- Option A: This is incorrect due to an error in the calculation.
- Option C: This is incorrect due to an error in the calculation.
- Option D: This is incorrect due to an error in the calculation.

Key Concepts:
- Photoelectric effect
- Kinetic energy
- Radius of circular path in a magnetic field

Memory Aids:
- Remember the formulas: $KE_{max} = \frac{hc}{\lambda} - \phi$ and $r = \frac{mv}{qB}$

Quick Tips:
- Convert all units to SI units before substituting into the formula.
- Be careful with the powers of 10.

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Analyzing +4 | -1

Assertion: Photoelectric effect is seen only when electron are bound.

Reason: Free electron cannot absorb whole energy of photon and consuming momentum and energy.

View answer and explanation

Correct option: A

If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion

Option A (If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion) is correct because the photoelectric effect requires bound electrons, and free electrons cannot absorb the entire photon energy and momentum.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion. The photoelectric effect requires bound electrons, and free electrons cannot absorb the entire photon energy and momentum.

Step-by-Step Solution:

Step 1: Analyze the assertion: The photoelectric effect is observed only when electrons are bound to a material. This is because a bound electron can transfer momentum to the lattice, allowing energy and momentum to be conserved. Therefore, the assertion is true.

Step 2: Analyze the reason: A free electron cannot absorb the entire energy of a photon while simultaneously conserving both energy and momentum. This is because a free electron cannot transfer momentum to another object. Therefore, the reason is true and correctly explains the assertion.

Why other options are wrong:
- Option B: If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion is incorrect because the reason correctly explains the assertion.
- Option C: If Assertions is true but the reason is false is incorrect because both the assertion and the reason are true.
- Option D: If both Assertion and reason are false is incorrect because both the assertion and the reason are true.

Key Concepts:
- Photoelectric effect
- Conservation of energy
- Conservation of momentum
- Bound electrons
- Free electrons

Memory Aids:
- Remember that bound electrons can transfer momentum to the lattice.

Quick Tips:
- Understand the role of bound electrons in the photoelectric effect.

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Understanding +4 | -1

The electrons are emitted in the photoelectric effect from a metal surface

View answer and explanation

Correct option: A

Only if the frequency of the incident radiation is above a certain threshold value

Option A (Only if the frequency of the incident radiation is above a certain threshold value) is correct because the photoelectric effect only occurs if the frequency of the incident light is greater than the threshold frequency of the metal.

Solution approach

Correct Answer + Core Reason: Only if the frequency of the incident radiation is above a certain threshold value. Photoelectric effect requires frequency above the threshold.

Step-by-Step Solution:

Step 1: Recall the concept of the threshold frequency in the photoelectric effect. The threshold frequency is the minimum frequency of light required to eject electrons from a metal surface.

Step 2: If the frequency of the incident light is below the threshold frequency, no electrons will be emitted, regardless of the intensity of the light.

Step 3: If the frequency is above the threshold, electrons will be emitted, and their kinetic energy will depend on the frequency of the light.

Why other options are wrong:
- Option B (Only if the temperature of the surface is high): While temperature can affect electron emission (thermionic emission), it's not the primary factor in the photoelectric effect.
- Option C (At a rate that is independent of the nature of the metal): The rate of electron emission depends on the metal's work function and the light's intensity.
- Option D (With a maximum velocity proportional to the frequency of the incident radiation): The maximum kinetic energy (and thus velocity) is related to the frequency, but not directly proportional. It's proportional to (frequency - threshold frequency).

Key Concepts:
- Photoelectric effect
- Threshold frequency
- Work function

Memory Aids:
- Remember that frequency is the key to initiating the photoelectric effect.

Quick Tips:
- Always consider the threshold frequency when analyzing photoelectric effect problems.

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Application +4 | -1

Find the de-Broglie wavelength for electrons which passed through a potential difference of 100 V ?

\left[ {h = 6.625 \times {{10}^{ - 34}}J - \sec } \right]

View answer and explanation

Correct option: A

1.228\mathop \mathrm{A}\limits^ \circ

Option A (

1.228\mathop \mathrm{A}\limits^ \circ

) is correct because it is calculated using the formula for de-Broglie wavelength:

\lambda = \frac{12.27}{\sqrt{V}} \mathop \mathrm{A}\limits^ \circ

.

Solution approach

Correct Answer + Core Reason: $1.228\mathop \mathrm{A}\limits^ \circ$ . The de-Broglie wavelength is calculated using the formula $\lambda = \frac{12.27}{\sqrt{V}} \mathop \mathrm{A}\limits^ \circ$ .

Step-by-Step Solution:

Step 1: Identify the potential difference $V = 100 \text{ V}$ .

Step 2: Use the formula for the de-Broglie wavelength of an electron accelerated through a potential difference $V$ : $\lambda = \frac{h}{\sqrt{2m_e eV}}$ , where $h$ is Planck's constant, $m_e$ is the mass of the electron, and $e$ is the elementary charge.

Step 3: Alternatively, use the simplified formula: $\lambda = \frac{12.27}{\sqrt{V}} \mathop \mathrm{A}\limits^ \circ$ .

Step 4: Substitute the value of $V$ into the simplified formula: $\lambda = \frac{12.27}{\sqrt{100}} \mathop \mathrm{A}\limits^ \circ$ .

Step 5: Calculate the wavelength: $\lambda = \frac{12.27}{10} \mathop \mathrm{A}\limits^ \circ = 1.227 \mathop \mathrm{A}\limits^ \circ \approx 1.228 \mathop \mathrm{A}\limits^ \circ$ .

Why other options are wrong:
- **Option B ( $2.228\mathop \mathrm{A}\limits^ \circ$ ): This value is incorrect and may result from an error in the calculation or using the wrong formula.
- Option C ( $3.228\mathop \mathrm{A}\limits^ \circ$ ): This value is incorrect and may result from an error in the calculation or using the wrong formula.
- Option D ( $4.228\mathop \mathrm{A}\limits^ \circ$ ): This value is incorrect and may result from an error in the calculation or using the wrong formula.

Key Concepts:
- de-Broglie wavelength
- Electron acceleration
- Potential difference

Memory Aids:**
- Remember the simplified formula: $\lambda = \frac{12.27}{\sqrt{V}} \mathop \mathrm{A}\limits^ \circ$ .

Quick Tips:
- Use the simplified formula for quick calculations.
- Ensure the potential difference is in volts.

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Understanding +4 | -1

e/m

ratio of anode rays produced in a discharge tube, depends on the

View answer and explanation

Correct option: A

Nature of the gas filled in the tube

Option A (Nature of the gas filled in the tube) is correct because the charge-to-mass ratio of anode rays depends on the ions formed from the gas used.

Solution approach

Correct Answer + Core Reason: Nature of the gas filled in the tube. The $e/m$ ratio of anode rays depends on the type of gas used in the discharge tube.

Step-by-Step Solution:

Step 1: Understand the formation of anode rays.

Step 2: Anode rays are positive ions formed when electrons collide with gas atoms in the discharge tube.

Step 3: The charge of the ion depends on the number of electrons removed from the gas atom.

Step 4: The mass of the ion depends on the mass of the gas atom.

Step 5: Therefore, the $e/m$ ratio depends on the nature of the gas.

Why other options are wrong:
- Option B: The material of the anode does not directly affect the $e/m$ ratio of the ions formed.
- Option C: The material of the cathode affects the electron emission, but not the $e/m$ ratio of the positive ions.
- Option D: Only the nature of the gas directly affects the $e/m$ ratio.

Key Concepts:
- Anode rays (positive ions)
- Discharge tube
- Charge-to-mass ratio

Memory Aids:
- Anode rays are made of positive ions from the gas.

Quick Tips:
- Remember that anode rays are positive ions formed from the gas in the tube.

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Understanding +4 | -1

If

V

be the accelerating voltage, then the maximum frequency of continuous

X

-rays is given by

View answer and explanation

Correct option: C

\frac{{eV}}{h}

Option C (

\frac{{eV}}{h}

) is correct because the maximum frequency of continuous X-rays is directly proportional to the accelerating voltage and inversely proportional to Planck's constant.

Solution approach

Correct Answer + Core Reason: $\frac{{eV}}{h}$ . The maximum frequency is directly proportional to the accelerating voltage.

Step-by-Step Solution:

Step 1: Recall the energy of an electron accelerated through a potential difference. The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $E = eV$ , where $e$ is the charge of the electron.

Step 2: Relate energy to frequency. The energy of a photon is given by $E = hf$ , where $h$ is Planck's constant and $f$ is the frequency of the photon.

Step 3: Equate the energies. In the case of continuous X-rays, the maximum energy of the X-ray photon is equal to the kinetic energy of the electron. Therefore, $eV = hf_{max}$ .

Step 4: Solve for the maximum frequency. Rearranging the equation, we get $f_{max} = \frac{{eV}}{h}$ .

Why other options are wrong:
- Option A: $\frac{{eh}}{V}$ . This is the inverse relationship and incorrect.
- Option B: $\frac{{hV}}{e}$ . This is also an incorrect relationship between frequency, voltage, and Planck's constant.
- Option D: $\frac{h}{{eV}}$ . This is the inverse of the correct answer and is incorrect.

Key Concepts:
- Kinetic energy of an electron in a potential difference
- Energy of a photon
- Continuous X-rays

Memory Aids:
- Remember $E = eV$ for electron energy and $E = hf$ for photon energy. Equate them to find the relationship between $V$ and $f$ .

Quick Tips:
- Always remember the relationship between energy, frequency, and voltage in the context of X-rays.

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