Photoelectric Effects and deBroglie Equation - NEET Physics Questions
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Photoelectric Effects and deBroglie Equation

Question 41: easy

Assertion (A): Electron from metal surface ejects only when light of particular wavelength will fall on surface.


Reason (R): Light shows wave nature.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Photoelectric emission occurs only when the incident light's wavelength (\lambda\) is below a specific threshold wavelength (\lambda_0\) (A is true). Light exhibits both wave-like (e.g., diffraction) and particle-like properties (R is true). However, the wave nature of light cannot explain the threshold effect for photoelectric emission. Thus, both A and R are true, but R is not the correct explanation for A.

Question 42: easy

Assertion (A): The ratio of wavelength in first transition of lyman series for H atom and He+ atom is exactly equal to four.


Reason (R): In all atoms electron revolve around fixed nucleus.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

The wavelength for hydrogen-like atoms is given by \(1/\lambda \propto Z^2\). For the first transition of the Lyman series (n=1 to n=2), \(lambda_H = 4/(3R)\) and \(lambda_{\text{He}^+} = 1/(3R)\). Their ratio \(lambda_H / \lambda_{\text{He}^+} = 4\). So (A) is true. Reason (R) is false as nuclei are not perfectly fixed; they exhibit recoil and vibration.

Question 43: easy

Assertion (A): When the speed of an electron increases its specific charge decreases.


Reason (R): Specific charge is the ratio of the mass to charge.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: relativistic mass \(m = m_0 \sqrt{1-v^2/c^2}\) increases with speed \(v\), so specific charge \(q/m\) decreases. Reason (R) is false because specific charge is defined as the ratio of charge to mass (\(q/m\)), not mass to charge.

Question 44: easy

Assertion (A): As we increase applied voltage on LED intensity of emitted light first increases then decreases.


Reason (R): We use LED in forward bias.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true in a nuanced sense; while intensity generally increases with voltage/current, at very high currents, efficiency can decrease due to thermal effects or Auger recombination (LED droop), leading to a peak and subsequent decrease in intensity. Reason (R) is true; LEDs are operated in forward bias. However, (R) does not explain the intensity variation described in (A).

Question 45: easy

The ground state energy of hydrogen atom is \(-13.6 \text{eV}\). The energy needed to ionize hydrogen atom from its second excited state will be

1. \(1.51 \text{eV}\)
2. \(3.4 \text{eV}\)
3. \(13.6 \text{eV}\)
4. \(6.8 \text{eV}\)
View Answer

Second excited state corresponds to \(n = 3\). The energy of this state is \(E_3 = -\frac{13.6}{3^2} = -1.51 \text{eV}\). Therefore, the energy required to ionize the electron from this level is \(+1.51 \text{eV}\).

Question 46: easy

The maximum kinetic energy of the emitted photoelectrons in photoelectric effects is independent of:

1. Frequency of incident radiation
2. Wavelength of incident radiation
3. Work function of material
4. Intensity of incident radiation
View Answer

The maximum kinetic energy of photoelectrons is given by Einstein's equation: \(K_{\text{max}} = h\nu - \Phi\). It depends on frequency, wavelength, and work function, but is independent of the light's intensity, which only affects the number of emitted photoelectrons.

Question 47: easy

The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:

1. 1.36 nm
2. 0.136 nm
3. 13.6 nm
4. 136 nm
View Answer

The de Broglie wavelength of an electron is \(\lambda = \frac{1.227}{\sqrt{V}} \text{nm}\). Substituting \(V = 81 \text{V}\), we get \(\lambda = \frac{1.227}{9} \approx 0.136 \text{nm}\).

Question 48: easy

Ultraviolet radiation of \(6.1 \text{eV}\) falls on an aluminium surface of work function \(4.2 \text{eV}\). The kinetic energy of the fastest electron emitted is

1. \(1.9 \times 10^{-19} \text{J}\)
2. \(1.6 \times 10^{-19} \text{J}\)
3. \(3.04 \times 10^{-19} \text{J}\)
4. \(2.5 \times 10^{-19} \text{J}\)
View Answer

According to Einstein's photoelectric equation, \(K_{max} = h\nu - \phi = 6.1 \text{eV} - 4.2 \text{eV} = 1.9 \text{eV}\). Converting to Joules: \(1.9 \times 1.6 \times 10^{-19} \text{J} = 3.04 \times 10^{-19} \text{J}\).

Question 49: easy

Given below are two statements:


Statement-I: The de Broglie wavelength (\(\lambda\)) associated with a moving particle is related to its momentum (\(p\)) as \(\lambda = \frac{h}{p}\).


Statement-II: Photons are electrically neutral and are not deflected by electric and magnetic fields.


In the light of above statements choose the most appropriate answer from the options given below.

1. Statement-I is correct but statement-II is incorrect
2. Statement-I is incorrect but statement-II is correct
3. Both statements-I and II are correct
4. Both statements-I and II are incorrect
View Answer

Statement-I is the basic de Broglie relationship \(\lambda = h/p\) which is correct. Statement-II is also correct as photons carry no charge and do not interact with electromagnetic fields.

Question 50: easy

In a photoelectric effect, a metal having threshold frequency ( f_0 ) is used. When it is irradiated with photons of frequency ( 2f_0 ), the stopping potential comes out to be ( V_0 ). If the same metal is irradiated with photons of frequency ( 3f_0 ), then the stopping potential will be equal to

1. \( 2V_0 \)
2. \( \frac{2}{3}V_0 \)
3. \( \frac{3}{2}V_0 \)
4. \( V_0 \)
View Answer

From Einstein's photoelectric equation, \( eV_0 = h(2f_0) - hf_0 = hf_0 \). For frequency ( 3f_0 ), \( eV' = h(3f_0) - hf_0 = 2hf_0 = 2(eV_0) \), which gives \( V' = 2V_0 \).