DUAL NATURE OF RADIATION AND MATTER MCQs for NEET UG PHYSICS — Medium, Page 3

61 Medium

Understanding +4 | -1

According to photon theory of light which of the following physical quantities associated with a photon do not/does not change as it collides with an electron in vacuum

View answer and explanation

Correct option: C

Speed only

Option C (Speed only) is correct because the speed of a photon in a vacuum is constant, while energy and momentum change during collision.

Solution approach

Correct Answer + Core Reason: Speed only. The speed of a photon in a vacuum is a constant, while its energy and momentum can change during collisions.

Step-by-Step Solution:

Step 1: Understand the properties of a photon.

Step 2: Recall that photons always travel at the speed of light in a vacuum.

Step 3: Consider the collision with an electron. The photon can transfer some of its energy and momentum to the electron.

Step 4: Since energy and momentum are transferred, they change. However, the photon still travels at the speed of light.

Step 5: Conclusion: Only the speed remains constant.

Why other options are wrong:
- Option A: Energy and momentum both change during the collision.
- Option B: Momentum changes, and speed remains constant.
- Option D: Energy changes, but speed remains constant.

Key Concepts:
- Photon theory of light
- Speed of light in vacuum
- Conservation of energy and momentum

Memory Aids:
- Photons always travel at the speed of light in a vacuum.

Quick Tips:
- Remember that the speed of light in a vacuum is constant for all photons.

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Remembering +4 | -1

The de-Broglie wavelength is proportional to

View answer and explanation

Correct option: C

\lambda \propto \frac{1}{p}

Option C is correct because the de Broglie wavelength is inversely proportional to the momentum (p) of the particle, as given by the formula λ = h/p.

Solution approach

Correct Answer + Core Reason: $\lambda \propto \frac{{1}}{{p}}$

Step-by-Step Solution:

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

Step 2: Analyze the relationship. From the formula, it is clear that the de Broglie wavelength is inversely proportional to the momentum (p).

Why other options are wrong:
- Option A: Wavelength is inversely proportional to velocity, but the question asks for proportionality to momentum directly. While $p=mv$ , the direct relationship is with p.
- Option B: Wavelength is inversely proportional to mass, but the question asks for proportionality to momentum directly. While $p=mv$ , the direct relationship is with p.
- Option D: Wavelength is inversely proportional to momentum, not directly proportional.

Key Concepts:
- de Broglie wavelength
- Momentum

Memory Aids:
- $\lambda = h/p$ (wavelength is inversely proportional to momentum).

Quick Tips:
- Remember the inverse relationship between de Broglie wavelength and momentum.

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Remembering +4 | -1

The de-Broglie wavelength associated with the particle of mass

m

moving with velocity

v

is

View answer and explanation

Correct option: A

h/mv

Option A is correct because the de Broglie wavelength is given by λ = h/mv, where h is Planck's constant, m is the mass, and v is the velocity.

Solution approach

Correct Answer + Core Reason: $h/mv$

Step-by-Step Solution:

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

Step 2: Relate momentum to mass and velocity. The momentum (p) of a particle is given by: $p = mv$ , where m is the mass and v is the velocity.

Step 3: Substitute the expression for momentum into the de Broglie wavelength formula. Substituting $p = mv$ into $\lambda = h/p$ gives: $\lambda = h/(mv)$ .

Why other options are wrong:
- Option B: This option has the variables inverted.
- Option C: This option has the variables in the wrong places.
- Option D: This option has the variables in the wrong places.

Key Concepts:
- de Broglie wavelength
- Momentum

Memory Aids:
- $\lambda = h/mv$ (de Broglie wavelength formula).

Quick Tips:
- Remember the relationship between momentum, mass, and velocity: $p = mv$ .

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Application +4 | -1

Light of wavelength 500 nm is incident on a metal surface with work function 2.28 eV. Maximum kinetic energy of emitted electron is

View answer and explanation

Correct option: C

0.2 eV

Option C (0.2 eV) is correct because the maximum kinetic energy is calculated using the formula

KE_{max} = E - \phi

, where

E = hc/\lambda

and

\phi

is the work function.

Solution approach

Correct Answer + Core Reason: 0.2 eV. The maximum kinetic energy is calculated using the formula $KE_{max} = E - \phi$ .

Step-by-Step Solution:

Step 1: Identify the given information: Wavelength $\lambda = 500 \mathrm{nm} = 500 \times 10^{-9} \mathrm{m}$ , Work function $\phi = 2.28 \mathrm{eV}$ .

Step 2: Calculate the energy of the incident photon: $E = \frac{hc}{\lambda}$ , where $h = 6.626 \times 10^{-34} \mathrm{Js}$ and $c = 3 \times 10^8 \mathrm{m/s}$ .

Step 3: $E = \frac{(6.626 \times 10^{-34} \mathrm{Js})(3 \times 10^8 \mathrm{m/s})}{500 \times 10^{-9} \mathrm{m}} = 3.9756 \times 10^{-19} \mathrm{J}$ .

Step 4: Convert the energy from Joules to eV: $E = \frac{3.9756 \times 10^{-19} \mathrm{J}}{1.602 \times 10^{-19} \mathrm{J/eV}} = 2.48 \mathrm{eV}$ .

Step 5: Use the photoelectric equation: $KE_{max} = E - \phi = 2.48 \mathrm{eV} - 2.28 \mathrm{eV} = 0.2 \mathrm{eV}$ .

Why other options are wrong:
- Option A: 0.1 eV is incorrect, likely from a miscalculation or incorrect formula.
- Option B: 0.5 eV is incorrect, likely from a miscalculation or incorrect formula.
- Option D: 0.6 eV is incorrect, likely from a miscalculation or incorrect formula.

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

Memory Aids:
- $E = hf = hc/\lambda$ (Energy = Planck's constant * frequency = Planck's constant * speed of light / wavelength)
- $KE_{max} = E - \phi$ (Kinetic Energy = Energy of photon - Work function)

Quick Tips:
- Remember to convert units to be consistent (e.g., Joules to eV).

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Application +4 | -1

How many photons are emitted by a laser source of

5 \times {10^{ - 3}}\;\mathrm{W}

operating at 632.2 nm in 2s?

View answer and explanation

Correct option: A

3.2 \times {10^{16}}

Option A (

3.2 \times {10^{16}}

) is correct because it is calculated by dividing the total energy emitted by the energy of a single photon.

Solution approach

Correct Answer + Core Reason: $3.2 \times {10^{16}}$ . The number of photons is calculated by dividing the total energy emitted by the energy of a single photon.

Step-by-Step Solution:

Step 1: Calculate the energy of a single photon: $E = hc/\lambda = (6.626 \times 10^{-34} \text{{ J s}} \times 3 \times 10^8 \text{{ m/s}}) / (632.2 \times 10^{-9} \text{{ m}}) \approx 3.14 \times 10^{-19} \text{{ J}}$ .

Step 2: Calculate the total energy emitted in 2 seconds: $E_{total} = P \times t = 5 \times 10^{-3} \text{{ W}} \times 2 \text{{ s}} = 10^{-2} \text{{ J}}$ .

Step 3: Calculate the number of photons emitted: $N = E_{total} / E = (10^{-2} \text{{ J}}) / (3.14 \times 10^{-19} \text{{ J}}) \approx 3.18 \times 10^{16}$ .

Step 4: Round to the closest option: $N \approx 3.2 \times 10^{16}$ .

Why other options are wrong:
- Option B: $1.6 \times {10^{16}}$ is incorrect due to calculation errors.
- Option C: $4 \times {10^{16}}$ is incorrect due to calculation errors.
- Option D: None of these is incorrect because option A is correct.

Key Concepts:
- Photon energy
- Power
- Energy of a photon

Memory Aids:
- $E = hc/\lambda$ and $E = Pt$ .

Quick Tips:
- Remember to use consistent units.

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Understanding +4 | -1

From the following, what charges can be present on oil drops in Millikan’s experiment

(Here

e

is the electronic charge)

View answer and explanation

Correct option: B

2e,\;1.6 \times {10^{ - 18}}C

Option B (

2e, 1.6 \times {10^{ - 18}}C

) is correct because the charge on the oil drops must be an integer multiple of the elementary charge, and

1.6 \times {10^{ - 18}}C = 10e

.

Solution approach

Correct Answer + Core Reason: $2e, 1.6 \times {10^{ - 18}}C$ because the charge must be an integer multiple of the elementary charge.

Step-by-Step Solution:

Step 1: Recall the principle of charge quantization in Millikan's experiment.

Step 2: The charge on any oil drop must be an integer multiple of the elementary charge, $e = 1.6 \times 10^{-19} C$ .

Step 3: Check each option to see if the charges are integer multiples of $e$ .

Step 4: Option B: $2e = 2 \times 1.6 \times 10^{-19} C = 3.2 \times 10^{-19} C$ . $1.6 \times 10^{-18} C = 10 \times 1.6 \times 10^{-19} C = 10e$ . Both are integer multiples of $e$ .

Why other options are wrong:
- Option A: Zero is a multiple of e, but the charge on an alpha particle is $2e$ , so it should be $3.2 \times 10^{-19} C$ .
- Option C: $1.6 \times 10^{-19} C = e$ , but $2.5e$ is not an integer multiple of $e$ .
- Option D: $1.5e$ is not an integer multiple of $e$ , and $e$ is fine.

Key Concepts:
- Millikan's oil drop experiment
- Charge quantization
- Elementary charge

Memory Aids:
- Charge is always a multiple of 'e'.

Quick Tips:
- Divide the given charge by $1.6 \times 10^{-19} C$ to check if it's an integer.

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Remembering +4 | -1

The de-Broglie wavelength associated with a particle moving with momentum

\left( p \right)

and mass

\left( m \right)

is

View answer and explanation

Correct option: A

\frac{h}{p}

Option A is correct because the de Broglie wavelength associated with a particle is given by λ = h/p, where h is Planck's constant and p is the momentum.

Solution approach

Correct Answer + Core Reason: $\frac{{h}}{{p}}$

Step-by-Step Solution:

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

Why other options are wrong:
- Option B: This option is incorrect as it includes mass in the denominator when it should only be momentum.
- Option C: This option is incorrect as it has momentum squared in the denominator.
- Option D: This option is incorrect as it has Planck's constant squared in the numerator and momentum squared in the denominator.

Key Concepts:
- de Broglie wavelength
- Momentum

Memory Aids:
- $\lambda = h/p$ (de Broglie wavelength formula).

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

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Understanding +4 | -1

In photoelectric effect if the intensity of light is doubled, then maximum kinetic energy of photoelectrons will become

View answer and explanation

Correct option: D

No change

Option D (No change) is correct because the maximum kinetic energy of photoelectrons depends on the frequency of light, not the intensity.

Solution approach

Correct Answer + Core Reason: No change. Kinetic energy depends on frequency, not intensity.

Step-by-Step Solution:

Step 1: Recall Einstein's photoelectric equation: $E_k = hv - \phi$ , where $E_k$ is the maximum 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 intensity of light is related to the number of photons, not the energy of each photon.

Step 3: Therefore, changing the intensity of light only changes the number of photoelectrons emitted, not their maximum kinetic energy.

Why other options are wrong:
- Option A (Double): Doubling the intensity doubles the number of photoelectrons, not their kinetic energy.
- Option B (Half): The kinetic energy is independent of the intensity.
- Option C (Four times): The kinetic energy is independent of the intensity.

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

Memory Aids:
- Intensity affects the number of electrons; frequency affects their kinetic energy.

Quick Tips:
- Distinguish between the effects of intensity and frequency in the photoelectric effect.

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Understanding +4 | -1

Electric field and magnetic field in Thomson mass spectrograph are applied

View answer and explanation

Correct option: D

Parallel simultaneously

Option D is correct because in Thomson's mass spectrograph, the electric and magnetic fields are applied parallel and simultaneously to deflect the charged particles.

Solution approach

Correct Answer + Core Reason: Parallel simultaneously

Step-by-Step Solution:

Step 1: Describe Thomson's mass spectrograph. Thomson's mass spectrograph uses electric and magnetic fields to deflect charged particles.

Step 2: Explain the arrangement of the fields. The electric and magnetic fields are applied parallel to each other and simultaneously.

Step 3: Explain the purpose of the fields. The parallel electric and magnetic fields cause the charged particles to be deflected in a way that depends on their charge-to-mass ratio.

Why other options are wrong:
- Option A: The fields are parallel, not perpendicular.
- Option B: The fields are applied simultaneously.
- Option C: The fields are parallel and applied simultaneously.

Key Concepts:
- Thomson's mass spectrograph
- Electric field
- Magnetic field

Memory Aids:
- Remember that the fields are parallel and simultaneous in Thomson's mass spectrograph.

Quick Tips:
- The parallel fields are key to the operation of Thomson's mass spectrograph.

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Analyzing +4 | -1

Assertion: Photo electric current increases if the distance between cathode and anode is increased.

Reason: Momentum of photon is directly proportional to its wavelength.

View answer and explanation

Correct option: D

If both Assertion and reason are false

Option D (If both Assertion and reason are false) is correct because both the assertion and the reason are incorrect statements.

Solution approach

Correct Answer + Core Reason: If both Assertion and reason are false. Both statements are incorrect.

Step-by-Step Solution:

Step 1: Analyze the assertion: Photoelectric current decreases if the distance between the cathode and anode is increased. This is because the collection efficiency of emitted electrons decreases with increased distance. So, the assertion is false.

Step 2: Analyze the reason: Momentum of a photon is inversely proportional to its wavelength ( $p = \frac{h}{\lambda}$ ). Therefore, the reason is false.

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

Key Concepts:
- Photoelectric effect
- Photoelectric current
- Momentum of a photon
- Wavelength of a photon

Memory Aids:
- Remember that increasing the distance between cathode and anode reduces the collection efficiency.
- Remember the relationship between momentum and wavelength: $p = \frac{h}{\lambda}$ .

Quick Tips:
- Carefully analyze both the assertion and the reason to determine if they are true.

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Remembering +4 | -1

\alpha -

particle contain –

View answer and explanation

Correct option: C

2 proton, 2 neutron only

Option C (2 proton, 2 neutron only) is correct because an alpha particle is the nucleus of a Helium atom, consisting of two protons and two neutrons.

Solution approach

Correct Answer + Core Reason: 2 proton, 2 neutron only. An alpha particle is a Helium nucleus.

Step-by-Step Solution:

Step 1: Recall the composition of an alpha particle. An alpha particle is equivalent to the nucleus of a helium atom.

Step 2: The helium nucleus consists of 2 protons and 2 neutrons.

Step 3: Therefore, an alpha particle contains 2 protons and 2 neutrons.

Why other options are wrong:
- Option A (2 proton, 2 neutron, 2 electron): This describes a neutral helium atom, not an alpha particle. Alpha particles are nuclei and do not have electrons.
- Option B (1 proton, 2 neutron, 1 electron): This is not the composition of an alpha particle or a helium atom.
- Option D (Single ionized Helium atom): A singly ionized helium atom has 2 protons, 2 neutrons, and 1 electron. An alpha particle has no electrons.

Key Concepts:
- Alpha particle composition
- Helium nucleus

Memory Aids:
- Alpha particle = Helium nucleus = 2 protons + 2 neutrons.

Quick Tips:
- Remember that alpha particles are positively charged because they lack electrons.

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Analyzing +4 | -1

Consider the two following statements I and II, and identify the correct choice given in the answers

III. In photovoltaic cells the photoelectric current produced is not proportional to the intensity of incident light.

IV. In gas filled photoemissive cells, the velocity of photoelectrons depends on the wavelength of the incident radiation.

View answer and explanation

Correct option: D

I is false but II is true

Option D (I is false but II is true) is correct because the photoelectric current is proportional to the intensity of light, and the velocity of photoelectrons in gas-filled cells does depend on the wavelength.

Solution approach

Correct Answer + Core Reason: I is false but II is true. Photoelectric current is proportional to intensity, and velocity in gas-filled cells depends on wavelength.

Step-by-Step Solution:

Step 1: Analyze Statement I: In photovoltaic cells, the photoelectric current produced is not proportional to the intensity of incident light. This statement is false. The photoelectric current is directly proportional to the intensity of the incident light, assuming the frequency of the light is above the threshold frequency.

Step 2: Analyze Statement II: In gas-filled photoemissive cells, the velocity of photoelectrons depends on the wavelength of the incident radiation. This statement is true. The energy of the photoelectrons, and thus their velocity, depends on the wavelength (or frequency) of the incident radiation, as described by the photoelectric effect equation.

Step 3: Determine the correct choice based on the analysis of Statements I and II. Since Statement I is false and Statement II is true, the correct choice is D.

Why other options are wrong:
- Option A: Both I and II are true: Statement I is false, so this option is wrong.
- Option B: Both I and II are false: Statement II is true, so this option is wrong.
- Option C: I is true but II is false: Both of these are incorrect.

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

Memory Aids:
- Photoelectric current is proportional to intensity.
- Energy of photoelectrons depends on wavelength.

Quick Tips:
- Remember the relationship between photoelectric current and light intensity.
- Recall the photoelectric effect equation and how it relates energy, wavelength, and velocity.

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Analyzing +4 | -1

Assertion: The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies.

Reason: The work function of the metal is its characteristics property.

View answer and explanation

Correct option: B

Assertion is correct, reason is correct; reason is not a correct explanation for assertion

Option B is correct because the assertion is true (photoelectrons have a range of kinetic energies), the reason is true (work function is a material property), but the reason doesn't explain the assertion.

Solution approach

Correct Answer + Core Reason: Assertion is correct, reason is correct; reason is not a correct explanation for assertion. The spread in kinetic energies is due to the varying depths from which electrons are emitted, not directly due to the work function being a characteristic property.

Step-by-Step Solution:

Step 1: Understand the photoelectric effect.

Step 2: Recall that the kinetic energy of photoelectrons is given by $KE = hf - \phi$ , where $hf$ is the energy of the incident photon and $\phi$ is the work function.

Step 3: Recognize that even with monochromatic light (fixed $hf$ ), electrons emitted from deeper within the metal lose some energy due to collisions before escaping, resulting in a range of kinetic energies.

Step 4: The work function is indeed a characteristic property of the metal, but it only determines the minimum energy required to eject an electron, not the spread in kinetic energies.

Why other options are wrong:
- Option A: The reason is not a correct explanation for the assertion. The work function being a characteristic property doesn't directly cause the spread in kinetic energies.
- Option C: The reason is correct, so this option is incorrect.
- Option D: The assertion is correct, so this option is incorrect.

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

Memory Aids:
- $KE = hf - \phi$ (Einstein's photoelectric equation)

Quick Tips:
- Remember that photoelectrons can lose energy before escaping the metal, leading to a range of kinetic energies.

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Analyzing +4 | -1

Assertion: In process of photoelectric emission, all emitted electrons do not have same kinetic energy.

Reason: If radiation falling on photosensitive surface of a metal consists of different wavelength then energy acquired by electrons absorbing photons of different wavelengths shall be different.

View answer and explanation

Correct option: B

Assertion is correct, reason is correct; reason is not a correct explanation for assertion

Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both statements are true, but the reason doesn't directly explain the assertion.

Solution approach

Correct Answer + Core Reason: Assertion is correct, reason is correct; reason is not a correct explanation for assertion. Both statements are true, but the reason doesn't directly explain the assertion.

Step-by-Step Solution:

Step 1: Analyze the assertion: In photoelectric emission, electrons are emitted with a range of kinetic energies due to various factors, such as the depth from which they are emitted and collisions within the material. Therefore, the assertion is correct.

Step 2: Analyze the reason: If the incident radiation consists of different wavelengths, the photons will have different energies ( $E = \frac{{hc}}{\lambda}$ ). Electrons absorbing these photons will acquire different amounts of energy. Therefore, the reason is also correct.

Step 3: Determine if the reason explains the assertion: While both statements are true, the reason doesn't fully explain why *all* emitted electrons don't have the same kinetic energy. Even with a single wavelength, electrons can lose energy before escaping the metal.

Why other options are wrong:
- Option A: Assertion is correct, reason is correct; reason is a correct explanation for assertion is incorrect because the reason doesn't fully explain the assertion.
- Option C: Assertion is correct, reason is incorrect is incorrect because both the assertion and the reason are correct.
- Option D: Assertion is incorrect, reason is correct is incorrect because the assertion is also correct.

Key Concepts:
- Photoelectric effect
- Kinetic energy of photoelectrons
- Wavelength and energy of photons

Memory Aids:
- Remember that electrons can lose energy through collisions before escaping the metal.

Quick Tips:
- Carefully analyze both the assertion and the reason to determine if they are true and if the reason explains the assertion.

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Understanding +4 | -1

The work function of a metal is

{W_0}

and

\lambda

is wavelength of incident radiation. There is no emission of photoelectron when

View answer and explanation

Correct option: D

\lambda > \frac{{hc}}{{{W_0}}}

Option D (

\lambda > \frac{{hc}}{{{W_0}}}

) is correct because for no photoemission to occur, the wavelength of the incident radiation must be greater than the threshold wavelength.

Solution approach

Correct Answer + Core Reason: $\lambda > \frac{{hc}}{{{W_0}}}$ . No emission occurs when the wavelength is greater than the threshold wavelength.

Step-by-Step Solution:

Step 1: Recall the photoelectric effect equation: $E = hv = \frac{{hc}}{\lambda}$ , where $E$ is the energy of the photon, $h$ is Planck's constant, $v$ is the frequency, $c$ is the speed of light, and $\lambda$ is the wavelength.

Step 2: The condition for photoelectric emission is that the energy of the photon must be greater than the work function ${W_0}$ : $E > {W_0}$ .

Step 3: Substitute $E = \frac{{hc}}{\lambda}$ into the inequality: $\frac{{hc}}{\lambda} > {W_0}$ .

Step 4: Rearrange the inequality to solve for $\lambda$ : $\lambda < \frac{{hc}}{{{W_0}}}$ .

Step 5: For no emission, the condition is the opposite: $\lambda > \frac{{hc}}{{{W_0}}}$ .

Why other options are wrong:
- **Option A ( $\lambda = \frac{{hc}}{{{W_0}}}$ ): This is the threshold wavelength, where emission is just about to occur.
- Option B ( $\lambda < \frac{{hc}}{{{W_0}}}$ ): This is the condition for photoelectric emission to occur, not for no emission.
- Option C ( $\lambda \le \frac{{hc}}{{{W_0}}}$ ): This also allows for emission to occur at the threshold wavelength.

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

Memory Aids:
- Remember that shorter wavelengths (higher energy) are needed for photoelectric emission.

Quick Tips:**
- Relate wavelength and energy using $E = \frac{{hc}}{\lambda}$ .

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Analyzing +4 | -1

While doing his experiment, Millikan one day observed the following charges on a single drop
(i)

6.563 \times {10^{ - 19}}C

(ii)

8.204 \times {10^{ - 19}}C

(iii)

11.50\; \times {10^{ - 19}}C\;

(iv)

13.13 \times {10^{ - 19}}C

(v)

16.48 \times {10^{ - 19}}C\;\;

(vi)

18.09 \times {10^{ - 19}}C

From this data the value of the elementary charge

\left( e \right)

was found to be

View answer and explanation

Correct option: A

1.641 \times {10^{ - 19}}C

Option A (

1.641 \times {10^{ - 19}}C

) is correct because it is the approximate elementary charge found by calculating the greatest common divisor of the observed charges.

Solution approach

Correct Answer + Core Reason: $1.641 \times {10^{ - 19}}C$ . The elementary charge is found by determining the greatest common divisor of the observed charges.

Step-by-Step Solution:

Step 1: The observed charges are multiples of the elementary charge $e$ . We need to find the best estimate for $e$ from the given data.

Step 2: Divide each charge by $10^{-19} C$ to simplify the numbers:
(i) 6.563
(ii) 8.204
(iii) 11.50
(iv) 13.13
(v) 16.48
(vi) 18.09

Step 3: Find the differences between consecutive charges:
(ii) - (i) = 8.204 - 6.563 = 1.641
(iii) - (ii) = 11.50 - 8.204 = 3.296
(iv) - (iii) = 13.13 - 11.50 = 1.63
(v) - (iv) = 16.48 - 13.13 = 3.35
(vi) - (v) = 18.09 - 16.48 = 1.61

Step 4: Look for a common factor among these differences. We see values around 1.6 and 3.3, where 3.3 is approximately twice 1.6.

Step 5: Divide each of the original charges by a value close to 1.6 to see if we get integers:
(i) 6.563 / 1.641 ≈ 4.00
(ii) 8.204 / 1.641 ≈ 5.00
(iii) 11.50 / 1.641 ≈ 7.01
(iv) 13.13 / 1.641 ≈ 8.00
(v) 16.48 / 1.641 ≈ 10.04
(vi) 18.09 / 1.641 ≈ 11.02

Step 6: The values are close to integers, so 1.641 is a good estimate for the elementary charge.

Step 7: Multiply by $10^{-19} C$ to get the elementary charge: $1.641 \times 10^{-19} C$ .

Why other options are wrong:
- **Option B ( $1.630 \times {10^{ - 19}}C$ ): While close, the value 1.641 gives better integer approximations when dividing the observed charges.
- Option C ( $1.648 \times {10^{ - 19}}C$ ): This value is slightly higher and does not give as good integer approximations.
- Option D ( $1.602 \times {10^{ - 19}}C$ ): This value is further away from the best estimate based on the data.

Key Concepts:
- Millikan's oil drop experiment
- Elementary charge
- Quantization of charge

Memory Aids:
- Charge is quantized in integer multiples of the elementary charge.

Quick Tips:**

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Application +4 | -1

The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of

4.14 \times {10^{ - 14}}

J is (Boltzmann constant
=1.38

\times {10^{ - 23}}\;\mathrm{J}{\mathrm{K}^{ - 1}})\mathrm{\;}

View answer and explanation

Correct option: A

2 \times {10^9}

K

Option A (

2 \times {10^9}

K) is correct because it is calculated using the formula

E = \frac{3}{2}kT

.

Solution approach

Correct Answer + Core Reason: $2 \times {10^9}$ K. The temperature is calculated using the average kinetic energy formula.

Step-by-Step Solution:

Step 1: Relate the energy to temperature using the formula: $E = \frac{3}{2}kT$

Step 2: Solve for temperature: $T = \frac{{2E}}{{3k}}$

Step 3: Substitute the given values: $T = \frac{{2 \times 4.14 \times 10^{-14} J}}{{3 \times 1.38 \times 10^{-23} JK^{-1}}}$

Step 4: Calculate the temperature: $T = \frac{{8.28 \times 10^{-14}}}{{4.14 \times 10^{-23}}} K$

Step 5: Simplify: $T = 2 \times 10^9 K$

Why other options are wrong:
- Option B: ${10^9}$ K is incorrect because it is not the correct calculation using the given values and the formula.
- Option C: 6 $\times {10^9}$ K is incorrect because it is not the correct calculation using the given values and the formula.
- Option D: $3 \times {10^9}$ K is incorrect because it is not the correct calculation using the given values and the formula.

Key Concepts:
- Average kinetic energy: $E = \frac{3}{2}kT$
- Boltzmann constant: $k = 1.38 \times 10^{-23} JK^{-1}$

Memory Aids:
- Remember the formula E = (3/2)kT.

Quick Tips:
- Ensure you use consistent units when calculating the temperature.

78 Medium

78 of 81Question permalink

Remembering +4 | -1

Which of the following is not the property of a cathode ray

View answer and explanation

Correct option: D

It does not deflect in electric field

Option D (It does not deflect in electric field) is correct because cathode rays are streams of electrons, which are negatively charged particles, and thus deflect in an electric field.

Solution approach

Correct Answer + Core Reason: It does not deflect in electric field. Cathode rays are streams of electrons and are deflected by electric fields.

Step-by-Step Solution:

Step 1: Recall that cathode rays are streams of electrons.

Step 2: Electrons are negatively charged particles.

Step 3: Charged particles are deflected by electric fields. Therefore, cathode rays are deflected by electric fields.

Final Answer: It does not deflect in electric field

Why other options are wrong:
- Option A (It casts shadow): This is incorrect because cathode rays travel in straight lines and can be blocked by objects, casting shadows.
- Option B (It produces heating effect): This is incorrect because when cathode rays strike a material, they transfer energy, causing a heating effect.
- Option C (It produces flurosence): This is incorrect because cathode rays can excite certain materials, causing them to fluoresce.

Key Concepts:
- Cathode rays
- Electrons
- Electric fields

Memory Aids:
- Remember that cathode rays are streams of electrons.

Quick Tips:
- Know the basic properties of cathode rays.

79 Medium

79 of 81Question permalink

Remembering +4 | -1

Positive rays were discovered by

View answer and explanation

Correct option: B

Goldstein

Option B (Goldstein) is correct because Eugen Goldstein is credited with the discovery of positive rays (also known as canal rays).

Solution approach

Correct Answer + Core Reason: Goldstein. Eugen Goldstein discovered positive rays.

Step-by-Step Solution:

Step 1: Recall the history of atomic physics and the discovery of subatomic particles.

Step 2: Eugen Goldstein is credited with the discovery of positive rays (canal rays) in 1886.

Final Answer: Goldstein

Why other options are wrong:
- Option A (Thomson): This is incorrect because J.J. Thomson is credited with the discovery of the electron.
- Option C (W. Crookes): This is incorrect because William Crookes is known for his work on cathode rays, but not the discovery of positive rays.
- Option D (Rutherford): This is incorrect because Ernest Rutherford is known for his work on alpha particles and the discovery of the atomic nucleus.

Key Concepts:
- Positive rays (canal rays)
- History of atomic physics

Memory Aids:
- Remember that Goldstein discovered positive rays.

Quick Tips:
- Know the key scientists and their contributions to atomic physics.

80 Medium

80 of 81Question permalink

Understanding +4 | -1

When ultraviolet radiation is incident on a surface, no photoelectrons are emitted. If a second beam causes photoelectrons to be ejected, it may consist of

View answer and explanation

Correct option: B

X-rays

Option B (X-rays) is correct because X-rays have higher energy and frequency than ultraviolet radiation, allowing them to eject photoelectrons when UV radiation fails.

Solution approach

Correct Answer + Core Reason: X-rays have higher energy than ultraviolet radiation and can cause photoelectrons to be ejected.

Step-by-Step Solution:

Step 1: Recall the relationship between energy, frequency, and wavelength of electromagnetic radiation.

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

Step 3: X-rays have a higher frequency (and therefore higher energy) than ultraviolet radiation. If UV radiation doesn't cause photoemission, a higher energy radiation like X-rays might.

Why other options are wrong:
- Option A: Infra-red waves have lower energy than ultraviolet radiation.
- Option C: Visible light rays have lower energy than ultraviolet radiation.
- Option D: Radio waves have much lower energy than ultraviolet radiation.

Key Concepts:
- Photoelectric effect
- Electromagnetic spectrum
- Energy of a photon

Memory Aids:
- Higher frequency = Higher energy.

Quick Tips:
- Remember the order of the electromagnetic spectrum (radio, microwave, infrared, visible, ultraviolet, X-ray, gamma ray) in terms of increasing frequency and energy.

81 Medium

81 of 81Question permalink

Application +4 | -1

Two identical metal plates shown photoelectric effect by a light of wavelength

\lambda \mathop A\limits^ \circ

falls on plate A and

{\lambda _B}

on plate

B\left( {{\lambda _A} = 2{\lambda _B}} \right).

The maximum kinetic energy is

View answer and explanation

Correct option: B

{K_A} < {K_B}/2

Option B (

{K_A} < {K_B}/2

) is correct because the kinetic energy is inversely proportional to the wavelength, and the wavelength of A is twice that of B.

Solution approach

Correct Answer + Core Reason: ${K_A} < {K_B}/2$ . The maximum kinetic energy is inversely proportional to the wavelength.

Step-by-Step Solution:

Step 1: Define the maximum kinetic energy in terms of the incident photon energy and the work function: $K = \frac{{hc}}{\lambda} - \phi$ , where $\phi$ is the work function.

Step 2: Write the kinetic energies for plates A and B:
$K_A = \frac{{hc}}{{\lambda_A}} - \phi$ and $K_B = \frac{{hc}}{{\lambda_B}} - \phi$

Step 3: Given that $\lambda_A = 2\lambda_B$ , substitute this into the equation for $K_A$ :
$K_A = \frac{{hc}}{{2\lambda_B}} - \phi$

Step 4: We want to compare $K_A$ and $K_B/2$ . Let's find an expression for $K_B/2$ :
$\frac{{K_B}}{2} = \frac{1}{2} \left( \frac{{hc}}{{\lambda_B}} - \phi \right) = \frac{{hc}}{{2\lambda_B}} - \frac{\phi}{2}$

Step 5: Compare $K_A$ and $K_B/2$ :
$K_A = \frac{{hc}}{{2\lambda_B}} - \phi$ and $\frac{{K_B}}{2} = \frac{{hc}}{{2\lambda_B}} - \frac{\phi}{2}$ \nSince $\phi > \frac{\phi}{2}$ , we have $K_A < \frac{{K_B}}{2}$ .

Why other options are wrong:
- Option A: $2{K_A} = {K_B}$ is incorrect because the relationship between the kinetic energies is not a direct multiple due to the work function.
- Option C: ${K_A} = 2{K_B}$ is incorrect because the kinetic energy is inversely proportional to the wavelength, and the wavelength of A is twice that of B, so $K_A$ should be smaller than $K_B$ .
- Option D: ${K_A} = {K_B}/2$ is incorrect because while the wavelength of A is twice that of B, the work function affects the kinetic energy, making the relationship not exactly half.

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

Memory Aids:
- $K = \frac{{hc}}{\lambda} - \phi$ (Kinetic energy equals photon energy minus work function)

Quick Tips:
- Remember to account for the work function when comparing kinetic energies.
- The shorter the wavelength, the higher the kinetic energy of the photoelectrons.

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