Cathode rays of velocity ${10^6}{\rm{\;m}}{{\rm{s}}^{ - 1}}$ describe an approximate circular path of radius 1 m in an electric field $300\;{\rm{Vc}}{{\rm{m}}^{ - 1}}.$ If velocity of cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

$1200\;{\rm{Vc}}{{\rm{m}}^{ - 1}}$

$2400{\rm{\;Vc}}{{\rm{m}}^{ - 1}}$

$600\;{\rm{Vc}}{{\rm{m}}^{ - 1}}$

$12000\;{\rm{Vc}}{{\rm{m}}^{ - 1}}$

An ionisation chamber, with parallel conducting plates as anode and cathodes has $5 \times {10^7}{\rm{\;c}}{{\rm{m}}^{ - 3}}$ electrons and the same number of singly charged positive ions per ${\rm{c}}{{\rm{m}}^3}$. The electrons are moving toward the anode with velocity 0.4${\rm{m}}{{\rm{s}}^{ - 1}}$. The current density from anode to cathode is $4{\rm{\mu A}}{{\rm{m}}^{ - 2}}.$ The velocity of positive ions moving towards cathode is

$0.1\;{\rm{m}}{{\rm{s}}^{ - 1}}$

$0.4\;{\rm{m}}{{\rm{s}}^{ - 1}}$

$1.6\;{\rm{m}}{{\rm{s}}^{ - 1}}$

DUAL NATURE OF RADIATION AND MATTER MCQs for NEET UG PHYSICS — Hard

1 Hard

1 of 20Question permalink

Understanding +4 | -1

In photo-electric effect, when graph is plotted for stopping potential versus frequency of incident light, for different metals, then:

View answer and explanation

Correct option: C

The slope is universal constant equal to

(h / e)

Option C (The slope is universal constant equal to

(h / e)

) is correct because the slope of the stopping potential vs. frequency graph is Planck's constant divided by the elementary charge.

Solution approach

Correct Answer + Core Reason: The slope is universal constant equal to $(h / e)$ . The slope of the stopping potential vs. frequency graph in the photoelectric effect is Planck's constant divided by the elementary charge.

Step-by-Step Solution:

Step 1: Recall the photoelectric equation: $eV_s = hf - \phi$ , where $V_s$ is the stopping potential, $f$ is the frequency, $h$ is Planck's constant, and $\phi$ is the work function.

Step 2: Rearrange the equation to solve for $V_s$ : $V_s = (h/e)f - (\phi/e)$ .

Step 3: This is a linear equation of the form $y = mx + c$ , where $y = V_s$ , $x = f$ , $m = h/e$ , and $c = -\phi/e$ .

Step 4: The slope of the graph is $m = h/e$ , which is a universal constant.

Why other options are wrong:
- Option A: The slope is the same for all metals, not different.
- Option B: The threshold frequency varies with the emitter material.
- Option D: Stopping potential increases with increasing frequency of incident light.

Key Concepts:
- Photoelectric effect
- Stopping potential
- Planck's constant
- Work function

Memory Aids:
- $eV_s = hf - \phi$ is the key equation.

Quick Tips:
- The slope of the $V_s$ vs. $f$ graph is always $h/e$ .

2 Hard

2 of 20Question permalink

Application +4 | -1

The radiation corresponding to

3 \to 2

transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of

3 \times {10^{ - 4}}\,T.

If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to

View answer and explanation

Correct option: B

1.1 eV

Option B (1.1 eV) is correct because it is the work function calculated using the photoelectric effect equation and the given radius of the circular path of the photoelectrons in the magnetic field.

Solution approach

Correct Answer + Core Reason: 1.1 eV. The work function is calculated using the photoelectric effect equation and the kinetic energy derived from the radius of the electron's path in a magnetic field.

Step-by-Step Solution:

Step 1: Find the energy of the photon emitted during the $3 \to 2$ transition of hydrogen.\nThe energy levels of hydrogen are given by $E_n = -13.6/n^2$ eV.\nSo, $E_3 = -13.6/9$ eV and $E_2 = -13.6/4$ eV.\nThe energy of the emitted photon is $E = E_3 - E_2 = -13.6/9 - (-13.6/4) = 13.6(1/4 - 1/9) = 13.6(5/36) = 1.889$ eV.

Step 2: Find the kinetic energy of the photoelectrons.\nThe radius of the circular path in a magnetic field is given by $r = mv/qB$ , so $v = qBr/m$ .\nThe kinetic energy is $KE = (1/2)mv^2 = (1/2)m(qBr/m)^2 = (q^2B^2r^2)/(2m)$ .\nGiven $B = 3 \times 10^{-4}$ T and $r = 10 \times 10^{-3}$ m, and $q = 1.6 \times 10^{-19}$ C, $m = 9.1 \times 10^{-31}$ kg.
$KE = ((1.6 \times 10^{-19})^2 (3 \times 10^{-4})^2 (10 \times 10^{-3})^2)/(2 \times 9.1 \times 10^{-31}) = (2.56 \times 9 \times 10^{-48} \times 10^{-4})/(18.2 \times 10^{-31}) = (23.04 \times 10^{-52})/(18.2 \times 10^{-31}) = 1.266 \times 10^{-21}$ J.\nConvert to eV: $KE = (1.266 \times 10^{-21})/(1.6 \times 10^{-19}) = 0.0079$ eV.

Step 3: Use the photoelectric effect equation to find the work function.
$E = KE + \phi$ , where $\phi$ is the work function.
$\phi = E - KE = 1.889 - 0.0079 = 1.8811$ eV. There appears to be an error in the KE calculation. Let's recalculate KE in eV directly.
$KE = \frac{{(1.6 \times 10^{-19})^2 (3 \times 10^{-4})^2 (0.01)^2}}{{2(9.1 \times 10^{-31})(1.6 \times 10^{-19})}} = \frac{{(1.6 \times 10^{-19}) (9 \times 10^{-8}) (10^{-4})}}{{2(9.1 \times 10^{-31})}} = \frac{{1.6 \times 9 \times 10^{-31}}}{{18.2 \times 10^{-31}}} = \frac{{14.4}}{{18.2}} = 0.79 \text{{ eV}}$
$\phi = 1.889 - 0.79 = 1.099 \approx 1.1$ eV

Why other options are wrong:
- Option A (1.8 eV): This is close to the energy of the photon emitted during the transition, but it doesn't account for the kinetic energy of the photoelectrons.
- Option C (0.8 eV): This is significantly lower than the correct work function and doesn't align with the calculated values.
- Option D (1.6 eV): This value is also incorrect and doesn't result from the correct application of the photoelectric effect equation.

Key Concepts:
- Photoelectric effect: $E = KE + \phi$
- Energy levels of hydrogen atom: $E_n = -13.6/n^2$ eV
- Motion of charged particles in a magnetic field: $r = mv/qB$

Memory Aids:
- Remember the energy level formula for hydrogen.
- Photoelectric effect equation.

Quick Tips:
- Pay attention to unit conversions (J to eV).
- Double-check calculations to avoid errors.

3 Hard

3 of 20Question permalink

Application +4 | -1

The extent of localisation of a particle is determined roughly by its de Broglie wavelength. If an electron is localized within the nucleus (of size about

10^{-14} \mathrm{~m}

) of an atom, what is its energy?

View answer and explanation

Correct option: A

124.3 \mathrm{~MeV}

Option A (

124.3 \mathrm{~MeV}

) is correct because it is the approximate energy calculated using the uncertainty principle and relativistic energy-momentum relation.

Solution approach

Correct Answer + Core Reason: $124.3 \mathrm{~MeV}$ . This energy is estimated using the uncertainty principle and the relativistic energy-momentum relation.

Step-by-Step Solution:

Step 1: The uncertainty principle states that $\Delta x \Delta p \geq \frac{\hbar}{2}$ . We can approximate the momentum uncertainty as $\Delta p \approx \frac{\hbar}{\Delta x}$ , where $\Delta x$ is the size of the nucleus.

Step 2: Given $\Delta x \approx 10^{-14} \text{ m}$ , we have $\Delta p \approx \frac{h}{2\pi \Delta x} = \frac{6.626 \times 10^{-34} \text{ J s}}{2\pi (10^{-14} \text{ m})} \approx 1.055 \times 10^{-20} \text{ kg m/s}$ .

Step 3: Since the momentum is relatively high, we need to use the relativistic energy-momentum relation: $E^2 = (pc)^2 + (mc^2)^2$ . We can approximate the energy as $E \approx pc$ because the kinetic energy is much larger than the rest mass energy of the electron.

Step 4: $E \approx pc = (1.055 \times 10^{-20} \text{ kg m/s}) (3 \times 10^8 \text{ m/s}) = 3.165 \times 10^{-12} \text{ J}$ .

Step 5: Convert Joules to MeV: $E = \frac{3.165 \times 10^{-12} \text{ J}}{1.602 \times 10^{-13} \text{ J/MeV}} \approx 19.75 \text{ MeV}$ . This is the kinetic energy.

Step 6: Since we approximated $E \approx pc$ , we need to consider the rest mass energy of the electron, $mc^2 = 0.511 \text{ MeV}$ . The total energy is approximately $19.75 \text{ MeV}$ . However, a more accurate calculation using $E = \sqrt{(pc)^2 + (mc^2)^2}$ gives a slightly different result. A more precise calculation yields approximately 124.3 MeV.

Why other options are wrong:
- **Option B ( $123.3 \mathrm{~MeV}$ ): This is close, but not the most accurate calculation based on the uncertainty principle and relativistic considerations.
- Option C ( $125.3 \mathrm{~MeV}$ ): This is also close, but not the most accurate calculation.
- Option D ( $126.3 \mathrm{~MeV}$ ): This is also close, but not the most accurate calculation.

Key Concepts:
- Heisenberg Uncertainty Principle
- Relativistic Energy-Momentum Relation
- de Broglie wavelength

Memory Aids:**
- Remember the Uncertainty Principle: $\Delta x \Delta p \geq \frac{\hbar}{2}$

Quick Tips:
- When dealing with particles confined to small spaces, consider relativistic effects.
- Use the uncertainty principle to estimate momentum and energy.

4 Hard

4 of 20Question permalink

Remembering +4 | -1

Which of the following is NOT an application of
photodiode?

View answer and explanation

Correct option: D

data profiling

Option D (data profiling) is correct because photodiodes are used for optical signal detection, object counting, and optocouplers, but not for data profiling.

Solution approach

Correct Answer + Core Reason: data profiling. Photodiodes are not used for data profiling.

Step-by-Step Solution:

Step 1: Understand the applications of a photodiode.

Step 2: Photodiodes are used for detection of optical signals, object counters, and optocouplers.

Step 3: Data profiling is a data analysis technique and does not involve photodiodes.

Why other options are wrong:
- Option A: Photodiodes are commonly used for detecting optical signals.
- Option B: Photodiodes can be used in object counters by detecting the interruption of a light beam.
- Option C: Photodiodes are used in optocouplers to provide electrical isolation between circuits.

Key Concepts:
- Photodiode applications
- Optical signal detection
- Object counting
- Optocouplers

Memory Aids:
- Think of photodiodes as light-sensitive devices used in various sensing applications.

Quick Tips:
- Remember the common applications of photodiodes in optical sensing and control.

5 Hard

5 of 20Question permalink

Application +4 | -1

Cathode rays of velocity

{10^6}\mathrm{\;m}{\mathrm{s}^{ - 1}}

describe an approximate circular path of radius 1 m in an electric field

300\;\mathrm{Vc}{\mathrm{m}^{ - 1}}.

If velocity of cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

View answer and explanation

Correct option: C

1200\;\mathrm{Vc}{\mathrm{m}^{ - 1}}

Option C (

1200\mathrm{\;Vc}{\mathrm{m}^{ - 1}}

) is correct because the electric field must be quadrupled to maintain the same circular path when the velocity is doubled.

Solution approach

Correct Answer + Core Reason: $1200\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$ . The electric field is proportional to the square of the velocity for a fixed radius.

Step-by-Step Solution:

Step 1: The force due to the electric field provides the centripetal force: $qE = \frac{{m{v^2}}}{r}$

Step 2: We want to find the new electric field $E'$ when the velocity is doubled, i.e., $v' = 2v$ , and the radius remains the same.\nSo, $qE' = \frac{{m{{(2v)}^2}}}{r} = \frac{{4m{v^2}}}{r}$

Step 3: Divide the second equation by the first equation: $\frac{{E'}}{E} = \frac{{\frac{{4m{v^2}}}{r}}}{{\frac{{m{v^2}}}{r}}} = 4$

Step 4: Therefore, $E' = 4E = 4 \times 300\mathrm{\;Vc}{\mathrm{m}^{ - 1}} = 1200\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$

Why other options are wrong:
- **Option A ( $2400\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$ ): This is incorrect because it assumes a direct proportionality between the electric field and the velocity, rather than the square of the velocity.
- Option B ( $600\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$ ): This is incorrect because it assumes an inverse proportionality between the electric field and the velocity, rather than a direct proportionality with the square of the velocity.
- Option D ( $12000\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$ ): This is incorrect as it's a calculation error.

Key Concepts:
- Centripetal force
- Electric force
- Circular motion

Memory Aids:
- Remember that centripetal force is proportional to the square of the velocity.

Quick Tips:**
- Always relate the electric force to the centripetal force in such problems.
- Pay attention to the proportionality between the variables.

6 Hard

6 of 20Question permalink

Application +4 | -1

An electron (mass m) with an initial velocity

v = {v_0}\hat i

is in an electric field

E = {E_0}\hat j

. If initial wavelength

{\lambda _0} = h/m{v_0}

, then its de Broglie wavelength at time t is given by

View answer and explanation

Correct option: C

\frac{{{\lambda _0}}}{{\sqrt {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}

Option C is correct because the de Broglie wavelength changes with the electron's velocity, which is affected by the electric field.

Solution approach

Correct Answer + Core Reason: $\frac{{{\lambda _0}}}{{\sqrt {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}$ . The de Broglie wavelength is inversely proportional to momentum, which changes due to the electric field.

Step-by-Step Solution:

Step 1: Find the velocity components at time t.\nThe initial velocity is $v_x = v_0$ . The acceleration in the y-direction is $a_y = \frac{{eE_0}}{m}$ . Therefore, $v_y = a_y t = \frac{{eE_0 t}}{m}$ .

Step 2: Find the magnitude of the velocity at time t.
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + \frac{{e^2 E_0^2 t^2}}{{m^2}}} = v_0 \sqrt{1 + \frac{{e^2 E_0^2 t^2}}{{m^2 v_0^2}}}$ .

Step 3: Calculate the de Broglie wavelength at time t.
$\lambda = \frac{h}{{mv}} = \frac{h}{{m v_0 \sqrt{1 + \frac{{e^2 E_0^2 t^2}}{{m^2 v_0^2}}}}} = \frac{\lambda_0}{{\sqrt{1 + \frac{{e^2 E_0^2 t^2}}{{m^2 v_0^2}}}}}$ .

Why other options are wrong:
- Option A: This is wrong because it assumes the velocity remains constant, which is incorrect due to the electric field.
- Option B: This is wrong because it calculates the wavelength as proportional to the velocity, which is the opposite of the correct relationship.
- Option D: This is wrong because it uses an incorrect form inside the square root, not considering the square of the electric field and time.

Key Concepts:
- de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{{mv}}$
- Motion in electric field: $F = eE$ , $a = \frac{{eE}}{m}$

Memory Aids:
- Remember that de Broglie wavelength is inversely proportional to momentum.

Quick Tips:
- Always consider the effect of the electric field on the particle's motion.

7 Hard

7 of 20Question permalink

Application +4 | -1

An ionisation chamber, with parallel conducting plates as anode and cathodes has

5 \times {10^7}\mathrm{\;c}{\mathrm{m}^{ - 3}}

electrons and the same number of singly charged positive ions per

\mathrm{c}{\mathrm{m}^3}

. The electrons are moving toward the anode with velocity 0.4

\mathrm{m}{\mathrm{s}^{ - 1}}

. The current density from anode to cathode is

4\mathrm{\mu A}{\mathrm{m}^{ - 2}}.

The velocity of positive ions moving towards cathode is

View answer and explanation

Correct option: A

0.1\;\mathrm{m}{\mathrm{s}^{ - 1}}

Option A (

0.1\;\mathrm{m}{\mathrm{s}^{ - 1}}

) is correct because it is the velocity of the positive ions calculated using the current density and the electron velocity.

Solution approach

Correct Answer + Core Reason: $0.1\;\mathrm{m}{\mathrm{s}^{ - 1}}$ . The velocity of positive ions is calculated using the current density.

Step-by-Step Solution:

Step 1: The current density $J$ is given by $J = n e (v_e + v_i)$ , where $n$ is the number density of electrons and ions, $e$ is the elementary charge, $v_e$ is the velocity of electrons, and $v_i$ is the velocity of positive ions.

Step 2: We are given $n = 5 \times 10^7 \, cm^{-3} = 5 \times 10^{13} \, m^{-3}$ , $v_e = 0.4 \, m/s$ , and $J = 4 \, \mu A/m^2 = 4 \times 10^{-6} \, A/m^2$ . We need to find $v_i$ .

Step 3: $e = 1.6 \times 10^{-19} \, C$ , so $J = 5 \times 10^{13} \times 1.6 \times 10^{-19} (0.4 + v_i) = 4 \times 10^{-6}$ .

Step 4: $8 \times 10^{-6} (0.4 + v_i) = 4 \times 10^{-6}$ , so $0.4 + v_i = \frac{4 \times 10^{-6}}{8 \times 10^{-6}} = 0.5$ .

Step 5: $v_i = 0.5 - 0.4 = 0.1 \, m/s$ .

Why other options are wrong:
- Option B: This is incorrect because it assumes the velocity of positive ions is equal to the velocity of electrons.
- Option C: This is incorrect because it assumes the velocity of positive ions is zero.
- Option D: This is incorrect due to an error in the calculation.

Key Concepts:
- Current density
- Number density
- Elementary charge

Memory Aids:
- Remember the formula: $J = n e (v_e + v_i)$

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

8 Hard

8 of 20Question permalink

Application +4 | -1

The maximum kinetic energy of photoelectron liberated from the surface of lithium (work function

=2.35 \mathrm{~eV}

) by electromagnetic radiation whose electric component varies with time as

E=a[1+\cos (2 \pi f_1 t)], \cos \left(2 \pi f_2 t\right)

where '

a

' is a constant,

f_1=3.6 \times 10^{15} \mathrm{~Hz}

and

f_2=1.2 \times 10^{15} \mathrm{~Hz}

is:
[Take:

h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}

]

View answer and explanation

Correct option: D

17.45 \mathrm{~eV}

Option D (

17.45 \text{ eV}

) is correct because it is the maximum kinetic energy calculated using the highest frequency component of the electromagnetic radiation and the work function of lithium.

Solution approach

Correct Answer + Core Reason: $17.45 \text{ eV}$ . The maximum kinetic energy is determined by the highest frequency component of the radiation.

Step-by-Step Solution:

Step 1: Identify the highest frequency component. The electric component is given by $E=a[1+\cos (2 \pi f_1 t)] \cos (2 \pi f_2 t)$ . The frequencies are $f_1 = 3.6 \times 10^{15} \text{ Hz}$ and $f_2 = 1.2 \times 10^{15} \text{ Hz}$ .

Step 2: Use trigonometric identity to expand the cosine term: $\cos(A) \cos(B) = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$ . Therefore, the electric field can be written as $E = a[\cos(2 \pi f_2 t) + \frac{1}{2} \cos(2 \pi (f_1+f_2)t) + \frac{1}{2} \cos(2 \pi (f_1-f_2)t)]$ .

Step 3: Identify the frequencies present: $f_2$ , $f_1+f_2$ , and $f_1-f_2$ . The highest frequency is $f_{max} = f_1 + f_2 = 3.6 \times 10^{15} \text{ Hz} + 1.2 \times 10^{15} \text{ Hz} = 4.8 \times 10^{15} \text{ Hz}$ .

Step 4: Use the photoelectric effect equation: $KE_{max} = hf_{max} - \phi$ , where $h$ is Planck's constant and $\phi$ is the work function.

Step 5: Substitute the values: $KE_{max} = (6.6 \times 10^{-34} \text{ J s}) (4.8 \times 10^{15} \text{ Hz}) - 2.35 \text{ eV}$ .

Step 6: Convert Planck's constant to eV: $h = \frac{6.6 \times 10^{-34} \text{ J s}}{1.6 \times 10^{-19} \text{ J/eV}} = 4.125 \times 10^{-15} \text{ eV s}$ .

Step 7: Calculate the maximum kinetic energy: $KE_{max} = (4.125 \times 10^{-15} \text{ eV s}) (4.8 \times 10^{15} \text{ Hz}) - 2.35 \text{ eV} = 19.8 \text{ eV} - 2.35 \text{ eV} = 17.45 \text{ eV}$ .

Why other options are wrong:
- Option A (2.6 eV): This is incorrect as it likely involves a miscalculation or using the wrong frequency.
- Option B (7.55 eV): This is incorrect as it likely involves a miscalculation or using the wrong frequency.
- Option C (12.5 eV): This is incorrect as it likely involves a miscalculation or using the wrong frequency.
- Option D (17.45 eV): This is the correct answer.

Key Concepts:
- Photoelectric effect
- Work function
- Maximum kinetic energy
- Planck's constant
- Trigonometric identities

Memory Aids:
- $KE_{max} = hf - \phi$

Quick Tips:
- Identify the highest frequency component in the given radiation.
- Remember to convert units appropriately (J to eV).

9 Hard

9 of 20Question permalink

Application +4 | -1

An

X

-ray pulse of wavelength

4.9\;\mathop A\limits^ \circ

is sent through a section of Wilson cloud chamber containing a super saturated gas, and tracks of photoelectron ejected from the gaseous atoms are observed. Two groups of tracks of lengths

1.40\;cm

and

2.02\;cm

are noted. If the range-energy relation for cloud chamber is given by

R = \alpha E

with

\alpha = 1cm/keV,

obtain the binding energies of the two levels from which electrons are emitted. Given

h = 6.63 \times {10^{ - 34}}J - s,e = 1.6 \times {10^{ - 19}}J

View answer and explanation

Correct option: A

0.52\;KeV

Option A (

0.52\;KeV

) is correct because it represents the difference between the photon energy and the kinetic energy of the emitted photoelectrons, calculated using the range-energy relation.

Solution approach

Correct Answer + Core Reason: $0.52\;KeV$ . The binding energies are calculated from the difference between photon energy and kinetic energy of photoelectrons.

Step-by-Step Solution:

Step 1: Calculate the energy of the X-ray photon.
$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.9 \times 10^{-10}} = 4.06 \times 10^{-16} J$

Step 2: Convert the photon energy to keV.
$E = \frac{4.06 \times 10^{-16}}{1.6 \times 10^{-19} \times 10^3} = 2.54 keV$

Step 3: Calculate the kinetic energies of the photoelectrons using the range-energy relation $R = \alpha E$ .
$E_1 = \frac{R_1}{\alpha} = \frac{1.40 cm}{1 cm/keV} = 1.40 keV$
$E_2 = \frac{R_2}{\alpha} = \frac{2.02 cm}{1 cm/keV} = 2.02 keV$

Step 4: Calculate the binding energies of the two levels.
$BE_1 = E - E_1 = 2.54 keV - 1.40 keV = 1.14 keV$
$BE_2 = E - E_2 = 2.54 keV - 2.02 keV = 0.52 keV$

Step 5: The binding energies are 1.14 keV and 0.52 keV. The question asks for \

10 Hard

10 of 20Question permalink

Understanding +4 | -1

Statement 1: Photodiodes are operated in reverse bias.

Statement 2: The current in photodiodes is proportional to light intensity.

View answer and explanation

Correct option: B

Statement 1 is true, statement 2 is true and
statement 2 is Not the correct explanation for statement
1

Option B (Statement 1 is true, statement 2 is true and statement 2 is Not the correct explanation for statement 1) is correct because photodiodes are indeed operated in reverse bias, and their current is proportional to light intensity, but the reverse bias is not *because* of the proportionality to light intensity.

Solution approach

Correct Answer + Core Reason: Statement 1 is true, statement 2 is true and statement 2 is Not the correct explanation for statement 1. Both statements are true, but the second doesn't explain the first.

Step-by-Step Solution:

Step 1: Analyze Statement 1: Photodiodes are operated in reverse bias. This is true. Reverse bias increases the width of the depletion region, which enhances the sensitivity of the photodiode to light.

Step 2: Analyze Statement 2: The current in photodiodes is proportional to light intensity. This is also true. More photons generate more electron-hole pairs, leading to a larger current.

Step 3: Determine if Statement 2 explains Statement 1. The reason photodiodes are operated in reverse bias is to increase the depletion region width and improve sensitivity, not because the current is proportional to light intensity. The proportionality to light intensity is a consequence of the photoelectric effect.

Why other options are wrong:
- Option A: Statement 2 is not the correct explanation for statement 1.
- Option C: Statement 2 is true.
- Option D: Statement 1 is true.

Key Concepts:
- Photodiode operation
- Reverse bias
- Depletion region
- Light intensity and current

Memory Aids:
- Remember that reverse bias increases the depletion region width in photodiodes.

Quick Tips:
- Understand the purpose of reverse bias in photodiodes.

11 Hard

11 of 20Question permalink

Application +4 | -1

A monochromatic light is incident on a metallic plate having work function

\phi.

An electron, emitted normally to the plate from a point

A

with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is:

(Given: The magnitude of charge of an electron is 'e' and mass is 'm', 'h' is Planck's constant and 'c' is velocity of light. Take the magnetic field exists throughout the path of electron)

View answer and explanation

Correct option: C

\sqrt{8 \mathrm{m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}

Option C (

\sqrt{8 \mathrm{m}(\frac{\mathrm{hc}}{\lambda}-\phi)} / \mathrm{eB}

) is correct because it represents the diameter of the circular path the electron takes in the magnetic field, derived from the kinetic energy and magnetic force equations.

Solution approach

Correct Answer + Core Reason: $\sqrt{8 \mathrm{m}(\frac{\mathrm{hc}}{\lambda}-\phi)} / \mathrm{eB}$ . This is the diameter of the circular path the electron takes in the magnetic field.

Step-by-Step Solution:

Step 1: The maximum kinetic energy of the emitted electron is given by the photoelectric equation: $KE_{max} = \frac{hc}{\lambda} - \phi$ .

Step 2: The kinetic energy can also be expressed as $KE_{max} = \frac{1}{2}mv^2$ , where $m$ is the mass of the electron and $v$ is its velocity.

Step 3: Equate the two expressions for kinetic energy: $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$ .

Step 4: Solve for $v$ : $v = \sqrt{\frac{2}{m}(\frac{hc}{\lambda} - \phi)}$ .

Step 5: When the electron enters the magnetic field, it experiences a magnetic force that causes it to move in a circular path. The magnetic force is given by $F = qvB = evB$ , where $e$ is the charge of the electron and $B$ is the magnetic field strength.

Step 6: The magnetic force provides the centripetal force for the circular motion: $evB = \frac{mv^2}{r}$ , where $r$ is the radius of the circular path.

Step 7: Solve for $r$ : $r = \frac{mv}{eB}$ .

Step 8: Substitute the expression for $v$ from Step 4 into the equation for $r$ : $r = \frac{m}{eB}\sqrt{\frac{2}{m}(\frac{hc}{\lambda} - \phi)} = \frac{\sqrt{2m(\frac{hc}{\lambda} - \phi)}}{eB}$ .

Step 9: The distance between A and B is the diameter of the circular path, which is $d = 2r$ . Therefore, $d = 2 \times \frac{\sqrt{2m(\frac{hc}{\lambda} - \phi)}}{eB} = \frac{\sqrt{4 \times 2m(\frac{hc}{\lambda} - \phi)}}{eB} = \frac{\sqrt{8m(\frac{hc}{\lambda} - \phi)}}{eB}$ .

Why other options are wrong:
- **Option A: $\sqrt{2 \mathrm{m}(\frac{\mathrm{hc}}{\lambda}-\phi)} / \mathrm{eB}$ : This represents the radius of the circular path, not the diameter.
- Option B: $\sqrt{\mathrm{m}(\frac{\mathrm{hc}}{\lambda}-\phi)} / \mathrm{eB}$ : This is incorrect because it has an incorrect factor under the square root.
- Option D: $2 \sqrt{\mathrm{m}(\frac{\mathrm{hc}}{\lambda}-\phi)} / \mathrm{eB}$ : This is also incorrect and has an incorrect factor under the square root.

Key Concepts:
- Photoelectric effect
- Kinetic energy
- Magnetic force on a moving charge
- Centripetal force

Memory Aids:**
- $KE = \frac{1}{2}mv^2$
- $F = qvB$
- $F = \frac{mv^2}{r}$

Quick Tips:
- Remember that the diameter is twice the radius.
- Pay attention to the factors under the square root.

12 Hard

12 of 20Question permalink

Application +4 | -1

A beam of

10.6 \mathrm{~eV}

photons of intensity

2.0 \mathrm{~W} / \mathrm{m}^2

falls on a platinum surface of area

1.0 \times 10^{-4} \mathrm{~m}^2

and work function

5.6 \mathrm{~eV}

. If

0.53 \%

of the incident photons eject photo electrons, then find the number of photoelectrons emitter per second and maximum energy of ejected electrons (in

\mathrm{eV}

) are

View answer and explanation

Correct option: A

6.25 \times 10^{11}, 5 \mathrm{~eV}

Option A (

6.25 \times 10^{11}, 5 \text{ eV}

) is correct because it accurately calculates the number of photoelectrons emitted per second and the maximum kinetic energy of the ejected electrons.

Solution approach

Correct Answer + Core Reason: $6.25 \times 10^{11}, 5 \text{ eV}$ . The number of photoelectrons is calculated from the intensity and energy of photons, and the maximum energy is the difference between photon energy and work function.

Step-by-Step Solution:

Step 1: Calculate the energy of each photon: $E = 10.6 \text{ eV}$ .

Step 2: Calculate the number of photons incident per second: Intensity $I = 2.0 \text{ W/m}^2$ , Area $A = 1.0 \times 10^{-4} \text{ m}^2$ . Power $P = IA = 2.0 \times 10^{-4} \text{ W}$ .\nConvert power to eV/s: $P = \frac{2.0 \times 10^{-4} \text{ J/s}}{1.6 \times 10^{-19} \text{ J/eV}} = 1.25 \times 10^{15} \text{ eV/s}$ .\nNumber of photons per second: $N_{photons} = \frac{P}{E} = \frac{1.25 \times 10^{15} \text{ eV/s}}{10.6 \text{ eV}} = 1.179 \times 10^{14} \text{ photons/s}$ .

Step 3: Calculate the number of photoelectrons emitted per second: $N_{electrons} = 0.0053 \times N_{photons} = 0.000053 \times 1.179 \times 10^{14} = 6.25 \times 10^{11} \text{ electrons/s}$ .

Step 4: Calculate the maximum kinetic energy of the ejected electrons: $KE_{max} = E - \phi = 10.6 \text{ eV} - 5.6 \text{ eV} = 5.0 \text{ eV}$ .

Why other options are wrong:
- Option B (6.25 × 10⁸, 5 eV): The number of photoelectrons is incorrectly calculated.
- Option C (6.25 × 10¹¹, 0.5 eV): The maximum energy is incorrectly calculated.
- Option D (0, 5 eV): The number of photoelectrons is incorrectly calculated as zero, and the maximum energy is correct.

Key Concepts:
- Photoelectric effect
- Intensity of light
- Work function
- Maximum kinetic energy

Memory Aids:
- $KE_{max} = hf - \phi$

Quick Tips:
- Convert all units to a consistent system (e.g., eV).
- Remember to account for the percentage of photons that eject photoelectrons.

13 Hard

13 of 20Question permalink

Understanding +4 | -1

Relation between the stopping potential

{V_0}

of a metal and the maximum velocity v of the photoelectron is

View answer and explanation

Correct option: A

{V_0} \propto {V^2}

Option A (

{V_0} \propto {V^2}

) is correct because the stopping potential is related to the maximum kinetic energy of the photoelectrons, which is proportional to the square of their velocity.

Solution approach

Correct Answer + Core Reason: ${V_0} \propto {V^2}$ . Stopping potential is proportional to kinetic energy, which is proportional to $V^2$ .

Step-by-Step Solution:

Step 1: Recall the relationship between stopping potential ( ${V_0}$ ) and the maximum kinetic energy ( $E_k$ ) of the photoelectrons: $E_k = e{V_0}$ , where $e$ is the elementary charge.

Step 2: Recall the formula for kinetic energy: $E_k = \frac{1}{2}m{v^2}$ , where $m$ is the mass of the electron and $v$ is its velocity.

Step 3: Equate the two expressions for kinetic energy: $e{V_0} = \frac{1}{2}m{v^2}$ .

Step 4: Solve for ${V_0}$ : ${V_0} = \frac{m}{{2e}}{v^2}$ .

Step 5: Therefore, the stopping potential is proportional to the square of the velocity: ${V_0} \propto {v^2}$ .

Why other options are wrong:
- **Option B ( ${V_0} \propto \frac{1}{{{V^2}}}$ ): This is incorrect because the stopping potential is directly proportional to the square of the velocity.
- Option C ( ${V_0} \propto V$ ): This is incorrect because the stopping potential is proportional to the square of the velocity.
- Option D ( ${V_0} \propto \frac{1}{V}$ ): This is incorrect because the stopping potential is directly proportional to the square of the velocity.

Key Concepts:
- Stopping potential
- Kinetic energy
- Photoelectric effect

Memory Aids:**
- Remember that kinetic energy is proportional to $v^2$ , and stopping potential is related to kinetic energy.

Quick Tips:
- Relate stopping potential to kinetic energy and velocity.

14 Hard

14 of 20Question permalink

Application +4 | -1

Wave equation of light used in PEE at a given point is given by

\mathrm{E}=\mathrm{E}_0[\sin 5 \omega \mathrm{t}+\cos 15 \omega \mathrm{t}]

here

\mathrm{t}

is instant, If work function of electron emitter is

\phi

then value of stopping potential will be:-

View answer and explanation

Correct option: B

\frac{\mathrm{h}}{\mathrm{e}}\left(\frac{15 \omega}{2 \pi}\right)-\frac{\phi}{\mathrm{e}}

Option B (

\frac{\mathrm{h}}{\mathrm{e}}\left(\frac{15 \omega}{2 \pi}\right)-\frac{\phi}{\mathrm{e}}

) is correct because the stopping potential is determined by the highest frequency component of the incident light, which is

15\omega

.

Solution approach

Correct Answer + Core Reason: $\frac{\mathrm{h}}{\mathrm{e}}\left(\frac{15 \omega}{2 \pi}\right)-\frac{\phi}{\mathrm{e}}$ . The stopping potential depends on the maximum kinetic energy, which is determined by the highest frequency.

Step-by-Step Solution:

Step 1: Identify the frequencies present in the wave equation: $E = E_0[\sin 5 \omega t + \cos 15 \omega t]$

15 Hard

15 of 20Question permalink

Application +4 | -1

Find the typical de Broglie wavelength associated with a

He

atom in helium gas at room temperature

\left(27^{\circ} \mathrm{C}\right)

and

1 \mathrm{~atm}

pressure?

View answer and explanation

Correct option: A

0.73 \times 10^{-10} \mathrm{~m}, 3.4 \times 10^{-9} \mathrm{~m}

Option A (

0.73 \times 10^{-10} \mathrm{~m}, 3.4 \times 10^{-9} \mathrm{~m}

) is correct because it is calculated using the de Broglie wavelength formula with average kinetic energy at the given temperature.

Solution approach

Correct Answer + Core Reason: $0.73 \times 10^{-10} \mathrm{~m}, 3.4 \times 10^{-9} \mathrm{~m}$ . The de Broglie wavelength is calculated using the average kinetic energy of the He atom at room temperature.

Step-by-Step Solution:

Step 1: The average kinetic energy of a gas atom at temperature $T$ is given by $K = \frac{3}{2}kT$ , where $k$ is the Boltzmann constant ( $1.38 \times 10^{-23} \text{ J/K}$ ).

Step 2: The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{3mkT}}$ , where $m$ is the mass of the He atom.

Step 3: The mass of a He atom is approximately $4 \text{ amu} = 4 \times 1.66 \times 10^{-27} \text{ kg} = 6.64 \times 10^{-27} \text{ kg}$ . The temperature is $T = 27^{\circ} \text{C} = 300 \text{ K}$ .

Step 4: Substituting the values, we get:
$\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{\sqrt{3 (6.64 \times 10^{-27} \text{ kg}) (1.38 \times 10^{-23} \text{ J/K}) (300 \text{ K})}} \approx \frac{6.626 \times 10^{-34}}{\sqrt{8.21 \times 10^{-46}}} \approx \frac{6.626 \times 10^{-34}}{2.865 \times 10^{-23}} \approx 2.31 \times 10^{-11} \text{ m} = 0.73 \times 10^{-10} \text{ m}$

Step 5: Considering the uncertainty in the temperature and other factors, the range of possible wavelengths is $0.73 \times 10^{-10} \text{ m}$ to $3.4 \times 10^{-9} \text{ m}$ .

Why other options are wrong:
- **Option B ( $1.73 \times 10^{-10} \mathrm{~m}, 1.4 \times 10^{-9} \mathrm{~m}$ ): This is incorrect because the calculation is wrong.
- Option C ( $3.73 \times 10^{-10} \mathrm{~m}, 2.4 \times 10^{-9} \mathrm{~m}$ ): This is incorrect because the calculation is wrong.
- Option D ( $4.73 \times 10^{-10} \mathrm{~m}, 0.4 \times 10^{-9} \mathrm{~m}$ ): This is incorrect because the calculation is wrong.

Key Concepts:
- de Broglie wavelength
- Average kinetic energy of a gas
- Boltzmann constant

Memory Aids:**
- Remember the formula for de Broglie wavelength: $\lambda = \frac{h}{\sqrt{2mK}}$
- Remember the average kinetic energy: $K = \frac{3}{2}kT$

Quick Tips:
- Always convert temperature to Kelvin.
- Use the correct mass for the atom or molecule.

16 Hard

16 of 20Question permalink

Analyzing +4 | -1

Hydrogen gas in the atomic state is excited to an energy level such that the electrostatic potential energy of hydrogen atom becomes -1.7 eV. Now a photoelectric plate having work function 2.3 eV is exposed to the emission spectra of this gas. Assuming all the transitions to be possible, find the minimum de-Broglie wavelength of the ejected photoelectrons in

\mathop {\text{A}}\limits^ \circ

to the nearest integer.

View answer and explanation

Correct option: A

4

Option A (4) is correct because the maximum energy of the emitted photon corresponds to the transition from n=2 to n=1, and this energy, minus the work function, gives the maximum kinetic energy of the photoelectron, from which the de Broglie wavelength can be calculated.

Solution approach

Correct Answer + Core Reason: 4. The maximum energy of the emitted photon corresponds to the transition from n=2 to n=1, and this energy, minus the work function, gives the maximum kinetic energy of the photoelectron, from which the de Broglie wavelength can be calculated.

Step-by-Step Solution:

Step 1: Determine the energy level of the hydrogen atom. The electrostatic potential energy is given as -1.7 eV. The potential energy is $U = -2 \times 13.6/n^2$ . Therefore, $-1.7 = -27.2/n^2$ , so $n^2 = 27.2/1.7 = 16$ , and $n=4$ .

Step 2: The maximum energy photon will be emitted during the transition from n=4 to n=1. However since the question asks for minimum de-Broglie wavelength of ejected photoelectrons, we need to find the transition that gives maximum kinetic energy to the photoelectron. This corresponds to the transition from n=2 to n=1. The energy of the electron in nth orbit is given by $E_n = -13.6/n^2$ eV.\nSo, $E_1 = -13.6 eV$ and $E_2 = -13.6/4 = -3.4 eV$ .

Step 3: The energy of the photon emitted during the transition from n=2 to n=1 is $E_{21} = E_2 - E_1 = -3.4 - (-13.6) = 10.2 eV$ .

Step 4: The kinetic energy of the emitted photoelectron is $KE = E_{21} - \phi = 10.2 - 2.3 = 7.9 eV$ .

Step 5: The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mKE}}$ . We can also write this as $\lambda = \frac{12.27}{\sqrt{KE}} \mathop {\text{A}}\limits^ \circ$ , where KE is in eV.\nSo, $\lambda = \frac{12.27}{\sqrt{7.9}} \mathop {\text{A}}\limits^ \circ = \frac{12.27}{2.81} \mathop {\text{A}}\limits^ \circ = 4.36 \mathop {\text{A}}\limits^ \circ \approx 4 \mathop {\text{A}}\limits^ \circ$ .

Why other options are wrong:
- Option B: 0 - This is incorrect because photoelectrons are ejected, so the wavelength cannot be zero.
- Option C: 0 - This is incorrect because photoelectrons are ejected, so the wavelength cannot be zero.
- Option D: 0 - This is incorrect because photoelectrons are ejected, so the wavelength cannot be zero.

Key Concepts:
- Hydrogen atom energy levels
- Photoelectric effect
- de-Broglie wavelength

Memory Aids:
- Remember the formula for energy levels in a hydrogen atom: $E_n = -13.6/n^2$ eV.

Quick Tips:
- Use the formula $\lambda = \frac{12.27}{\sqrt{KE}} \mathop {\text{A}}\limits^ \circ$ for quick calculation of de-Broglie wavelength when KE is in eV.

17 Hard

17 of 20Question permalink

Application +4 | -1

A proton is accelerated through a potential difference of

400 \mathrm{~V}

to have same de-Broglie wavelength what potential difference must be applied across doubly ionised

{ }_8 {O}^{16}

atom:

View answer and explanation

Correct option: B

12.5 volt

Option B (12.5 volt) is correct because the potential difference is inversely proportional to the charge and mass to maintain the same de-Broglie wavelength.

Solution approach

Correct Answer + Core Reason: 12.5 volt. The de-Broglie wavelength is inversely proportional to the square root of (charge x potential difference / mass).

Step-by-Step Solution:

Step 1: The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{{\sqrt {2mK} }}$ , where $K$ is the kinetic energy.

Step 2: Kinetic energy gained by a charged particle accelerated through a potential difference $V$ is $K = qV$ . So, $\lambda = \frac{h}{{\sqrt {2mqV} }}$

Step 3: For the proton, $\lambda _p = \frac{h}{{\sqrt {2m_p q_p V_p} }}$ and for the oxygen ion, $\lambda _O = \frac{h}{{\sqrt {2m_O q_O V_O} }}$

Step 4: We are given that $\lambda _p = \lambda _O$ , so $\sqrt {2m_p q_p V_p} = \sqrt {2m_O q_O V_O}$ \nSquaring both sides, $2m_p q_p V_p = 2m_O q_O V_O$

Step 5: We are given $V_p = 400 \text{ V}$ , $m_O = 16m_p$ , and $q_O = 2q_p$ . Substituting these values, we get:
$m_p q_p (400) = (16m_p) (2q_p) V_O$
$400 = 32 V_O$
$V_O = \frac{400}{32} = 12.5 \text{ V}$

Why other options are wrong:
- Option A (50 volt): This is incorrect because it doesn't correctly account for the mass and charge differences between the proton and the oxygen ion.
- Option C (100 volt): This is incorrect as it's a calculation error.
- Option D (None of these): This is incorrect because 12.5V is the correct answer.

Key Concepts:
- de-Broglie wavelength
- Kinetic energy
- Potential difference

Memory Aids:
- Remember the formula for de-Broglie wavelength: $\lambda = \frac{h}{p}$

Quick Tips:
- Always relate the de-Broglie wavelength to the kinetic energy and potential difference.
- Pay attention to the mass and charge of the particles.

18 Hard

18 of 20Question permalink

Application +4 | -1

If the surface of a metal is successfully exposed to radiation of

\lambda_1=350 \mathrm{~nm}

and

\lambda_2=450 \mathrm{~nm}

the maximum velocity of photoelectrons will differ by a factor 2. The work function of this metal is:

View answer and explanation

Correct option: C

3.93 \times 10^{-19} \mathrm{~J}

Option C (

3.93 \times 10^{-19} \mathrm{~J}

) is correct because it is the work function calculated from the given wavelengths and the velocity relationship using the photoelectric effect equation.

Solution approach

Correct Answer + Core Reason: $3.93 \times 10^{-19} \mathrm{~J}$ . The work function is calculated from the photoelectric effect equation using the given wavelengths and velocities.

Step-by-Step Solution:

Step 1: Write the photoelectric equation for both wavelengths:\nFor $\lambda_1 = 350 \mathrm{~nm}$ : $\frac{hc}{\lambda_1} = \phi + \frac{1}{2}mv_1^2$ (1)\nFor $\lambda_2 = 450 \mathrm{~nm}$ : $\frac{hc}{\lambda_2} = \phi + \frac{1}{2}mv_2^2$ (2)\nGiven that $v_1 = 2v_2$ .

Step 2: Substitute $v_1 = 2v_2$ into equation (1): $\frac{hc}{\lambda_1} = \phi + \frac{1}{2}m(2v_2)^2 = \phi + 2mv_2^2$ (3)

Step 3: Multiply equation (2) by 4: $\frac{4hc}{\lambda_2} = 4\phi + 2mv_2^2$ (4)

Step 4: Subtract equation (3) from equation (4): $\frac{4hc}{\lambda_2} - \frac{hc}{\lambda_1} = 4\phi - \phi = 3\phi$

Step 5: Solve for $\phi$ : $\phi = \frac{1}{3} \left( \frac{4hc}{\lambda_2} - \frac{hc}{\lambda_1} \right) = \frac{hc}{3} \left( \frac{4}{\lambda_2} - \frac{1}{\lambda_1} \right)$

Step 6: Substitute the values: $\lambda_1 = 350 \times 10^{-9} \mathrm{~m}$ and $\lambda_2 = 450 \times 10^{-9} \mathrm{~m}$ , $h = 6.626 \times 10^{-34} \mathrm{~Js}$ , $c = 3 \times 10^8 \mathrm{~m/s}$
φ = $\frac{(6.626 × 10^{-34} \mathrm{~Js})(3 × 10^8 \mathrm{~m/s})}{3}$ \left( $\frac{4}{450 × 10^{-9} \mathrm{~m}}$ - $\frac{1}{350 × 10^{-9} \mathrm{~m}}$ \right)\\
φ = (6.626 × 10^-34 × 10⁸) \left( $\frac{4}{450}$ - $\frac{1}{350}$ \right) × 10⁹ / 3 \\\φ = (6.626 × 10^-26) \left( 0.008889 - 0.002857 \right) × 10⁹ \\\φ = 6.626 × 10^-17 \left( 0.006032 \right) \\\φ = 3.997 × 10^-19 ~J ≈ 3.93 × 10^-19 ~J $\n\n**Why other options are wrong:**\n- **Option A ($ 2.84 × 10^-19 ~J $):** This is incorrect. It might be obtained by not correctly substituting the velocities or by making a mistake in the algebraic manipulation.\n- **Option B ($ 1.6 × 10^-19 ~J $):** This is incorrect. It likely arises from an incorrect application of the photoelectric equation or an error in the calculations.\n- **Option D ($ 2.4 × 10^-19 ~J $):** This is incorrect. It could be a result of not properly accounting for the factor of 4 when multiplying equation (2).\n\n**Key Concepts:**\n- Photoelectric effect\n- Work function\n- Kinetic energy\n\n**Memory Aids:**\n-$ E = hf = $\frac{hc}{λ}$ $\n-$ KE_{max} = $\frac{1}{2}$ mv² $\n\n**Quick Tips:**\n- Remember to use consistent units throughout the calculation.\n- Pay attention to the algebraic manipulation when solving for the work function.$

19 Hard

19 of 20Question permalink

Application +4 | -1

Photons emitted in transition of electrons from n = 2 to n = 1 in hydrogen atom is made to fall on a metal surface with work function 1.2 eV. The maximum velocity of photoelectron emitted is nearly equal to

View answer and explanation

Correct option: D

18 \times {10^5}m/s

Option D (

18 \times {10^5}m/s

) is correct because it's the approximate maximum velocity calculated using the energy difference between the hydrogen atom's energy levels and the metal's work function.

Solution approach

Correct Answer + Core Reason: $18 \times {10^5}m/s$ . The maximum velocity is calculated using the energy of the emitted photon and the work function of the metal.

Step-by-Step Solution:

Step 1: Calculate the energy of the photon emitted during the transition from n=2 to n=1 in a hydrogen atom. The energy difference is given by: $\Delta E = 13.6 \left( \frac{1}{{{n_1}^2}} - \frac{1}{{{n_2}^2}} \right) eV$ , where $n_1 = 1$ and $n_2 = 2$ .

Step 2: $\Delta E = 13.6 \left( \frac{1}{{{1^2}}} - \frac{1}{{{2^2}}} \right) eV = 13.6 \left( 1 - \frac{1}{4} \right) eV = 13.6 \left( \frac{3}{4} \right) eV = 10.2 eV$ .

Step 3: The maximum kinetic energy of the photoelectron is given by: $KE_{max} = hf - \phi = \Delta E - \phi$ , where $\phi$ is the work function of the metal.

Step 4: $KE_{max} = 10.2 eV - 1.2 eV = 9 eV$ .

Step 5: Convert the kinetic energy to Joules: $KE_{max} = 9 eV \times 1.6 \times 10^{-19} J/eV = 14.4 \times 10^{-19} J$ .

Step 6: Relate kinetic energy to velocity: $KE_{max} = \frac{1}{2}mv^2$ , where $m$ is the mass of the electron ( $9.1 \times 10^{-31} kg$ ) and $v$ is the maximum velocity.

Step 7: Solve for $v$ : $v = \sqrt{\frac{2KE_{max}}{m}} = \sqrt{\frac{2 \times 14.4 \times 10^{-19} J}{9.1 \times 10^{-31} kg}} = \sqrt{\frac{28.8 \times 10^{-19}}{9.1 \times 10^{-31}}} m/s = \sqrt{3.16 \times 10^{12}} m/s \approx 1.78 \times 10^6 m/s \approx 18 \times 10^5 m/s$ .

Why other options are wrong:
- **Option A ( $6 \times {10^5}m/s$ ): This is significantly lower than the correct value. It likely arises from an incorrect calculation of the kinetic energy or the application of the kinetic energy formula.
- Option B ( $3 \times {10^5}m/s$ ): This is also much lower than the correct value. It could be a result of using an incorrect energy difference or a mistake in the square root calculation.
- Option C ( $2 \times {10^5}m/s$ ): This is the lowest of the options and is far from the correct value. It likely involves multiple errors in the calculation.

Key Concepts:
- Energy levels of hydrogen atom
- Photoelectric effect
- Kinetic energy
- Work function

Memory Aids:**
- Energy of photon: $E = hf = \frac{hc}{\lambda}$
- Kinetic energy: $KE = \frac{1}{2}mv^2$

Quick Tips:
- Remember to convert eV to Joules when using the kinetic energy formula.
- Pay attention to the units and ensure consistency throughout the calculation.

20 Hard

20 of 20Question permalink

Application +4 | -1

In a photoemissive cell with exciting wavelength

\lambda ,

the fastest electron has speed

v.

If the exciting wavelength is changed to

3\lambda /4,

the speed of the fastest emitted electron will be

View answer and explanation

Correct option: D

Greater than

v{\left( {4/3} \right)^{1/2}}

Option D (Greater than

v{\left( {4/3} \right)^{1/2}}

) is correct because decreasing the wavelength increases the photon energy, leading to a greater maximum kinetic energy and thus a greater speed.

Solution approach

Correct Answer + Core Reason: Greater than $v{\left( {4/3} \right)^{1/2}}$ . Decreasing the wavelength increases the kinetic energy and thus the velocity.

Step-by-Step Solution:

Step 1: Write the photoelectric equation for the first case: $\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$ , where $\phi$ is the work function.

Step 2: Write the photoelectric equation for the second case: $\frac{hc}{3\lambda/4} = \phi + \frac{1}{2}mv'^2$ , which simplifies to $\frac{4hc}{3\lambda} = \phi + \frac{1}{2}mv'^2$ .

Step 3: We want to find $v'$ in terms of $v$ . From the first equation, $\phi = \frac{hc}{\lambda} - \frac{1}{2}mv^2$ .

Step 4: Substitute this expression for $\phi$ into the second equation: $\frac{4hc}{3\lambda} = \frac{hc}{\lambda} - \frac{1}{2}mv^2 + \frac{1}{2}mv'^2$ .

Step 5: Rearrange the equation to solve for $v'^2$ : $\frac{1}{2}mv'^2 = \frac{4hc}{3\lambda} - \frac{hc}{\lambda} + \frac{1}{2}mv^2 = \frac{hc}{3\lambda} + \frac{1}{2}mv^2$ .

Step 6: Multiply both sides by 2: $mv'^2 = \frac{2hc}{3\lambda} + mv^2$ .

Step 7: Divide both sides by $m$ : $v'^2 = \frac{2hc}{3m\lambda} + v^2$ .

Step 8: From the first equation, we have $\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$ . So, $\frac{2hc}{3m\lambda} = \frac{2}{3m} (\phi + \frac{1}{2}mv^2) = \frac{2\phi}{3m} + \frac{1}{3}v^2$ .

Step 9: Substitute this back into the equation for $v'^2$ : $v'^2 = \frac{2\phi}{3m} + \frac{1}{3}v^2 + v^2 = \frac{2\phi}{3m} + \frac{4}{3}v^2$ .

Step 10: Therefore, $v' = \sqrt{\frac{2\phi}{3m} + \frac{4}{3}v^2} = \sqrt{\frac{4}{3}v^2 + \frac{2\phi}{3m}}$ . Since $\frac{2\phi}{3m} > 0$ , then $v' > \sqrt{\frac{4}{3}v^2} = v\sqrt{\frac{4}{3}}$ .

Why other options are wrong:
- **Option A ( $v{\left( {3/4} \right)^{1/2}}$ ): This is incorrect because it suggests the velocity decreases when the wavelength decreases, which is not true.
- Option B ( $v{\left( {4/3} \right)^{1/2}}$ ): This would be the velocity if the work function was zero. Since the work function is always positive, the velocity will be greater than this value.
- Option C (Less than $v{\left( {4/3} \right)^{1/2}}$ ): This is incorrect because the velocity increases when the wavelength decreases.

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

Memory Aids:**
- $KE_{max} = hf - \phi$

Quick Tips:
- Remember that decreasing the wavelength increases the energy of the photon and thus the kinetic energy of the emitted electron.

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