JEE Main Dual Nature of Radiation and Matter PYQ

56. (AIEEE 2006 )

The anode voltage of a photocell is kept fixed. The wavelength $λ$ of the light falling on the cathode is gradually changed. The plate current $I$ of the photocell varies as follows

(A) AIEEE 2006 Physics - Dual Nature of Radiation Question 128 English Option 1

(B) AIEEE 2006 Physics - Dual Nature of Radiation Question 128 English Option 2

(C) AIEEE 2006 Physics - Dual Nature of Radiation Question 128 English Option 3

(D) AIEEE 2006 Physics - Dual Nature of Radiation Question 128 English Option 4

Correct answer is option B

As $λ$ decreases, $v$ increases and hence the speed of photo electron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.

57. (AIEEE 2004 )

The work function of a substance is 4.0eV The longest wavelength of light that can cause photo-electron emission from this substance is approximately.

(A) 310nm

(B) 400nm

(C) 540nm

(D) 220nm

Correct answer is A

For the longest wavelength to emit photo electron

$\frac{h c}{λ} = ϕ \Rightarrow λ = \frac{h c}{ϕ}$

$\Rightarrow λ = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{40 \times 1.6 \times 10^{- 16}} = 310 n m$

58. (AIEEE 2004 )

According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal V_s the frequency, of the incident radiation gives as straight the whose slope

(A) depends both on the intensity of the radiation and the metal used

(B) depends on the intensity of the radiation

(C) depends on the nature of the metal used

(D) is the same for the all metals and independent of the intensity of the radiation

Correct answer is D

Einstein's photoelectric equation is given by

$K_{max} = h ν - ϕ,$

where $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons, $h$ is Planck's constant, $ν$ is the frequency of the incident radiation, and $ϕ$ is the work function of the metal (the minimum energy needed to eject an electron).

If you plot $K_{max}$ against $ν$ , you will get a straight line with slope $h$ (Planck's constant) and y-intercept $- ϕ$ (the negative of the work function). The slope of the line (which is $h$ ) does not depend on the intensity of the radiation or the type of metal used. Instead, it is a universal constant.

Therefore, the correct answer is:

Option D: The slope is the same for all metals and independent of the intensity of the radiation.

59. (AIEEE 2003 )

Two identical photo-cathodes receive light of frequencies $f_{1}$ and $f_{2}$ . If the velocities of the photo electrons (of mass $m$ ) coming out are respectively $v_{1}$ and $v_{2},$ then

(A) $v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

(B) $v_{1} + v_{2} = {[\frac{2 h}{m} (f_{1} + f_{2})]}^{1 / 2}$

(C) $v_{1}^{2} + v_{2}^{2} = \frac{2 h}{m} (f_{1} + f_{2})$

(D) $v_{1} - v_{2} = {[\frac{2 h}{m} (f_{1} - f_{2})]}^{1 / 2}$

Correct answer is option A

For one photo cathode

$h f_{1} - W = \frac{1}{2} m v_{1}^{2} . . . (i)$

For another photo cathode

$h f_{2} - W = \frac{1}{2} m v_{2}^{2} . . . (i i)$

Subtracting $(i i)$ from $(i)$ we get

$(h f_{1} - W) - (h f_{2} - W) = \frac{1}{2} m v_{1}^{2} - \frac{1}{2} m v_{2}^{2}$

$∴$ $h (f_{1} - f_{2}) = \frac{m}{2} (v_{1}^{2} - v_{2}^{2})$

$∴$ $v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

60. (AIEEE 2002 )

Sodium and copper have work functions 2.3eV and 4.5eV respectively. Then the ratio of the wavelengths is nearest to

(A) 1 : 2

(B) 4 : 1

(C) 2 : 1

(D) 1 : 4

Correct answer is C

We know that work function is the energy required and energy $E = h υ$

$∴$ $\frac{E_{N A}}{E_{C u}} = \frac{h υ_{N a}}{h υ_{c u}} = \frac{λ_{c u}}{λ_{N a}}$

$[$ as $υ \propto \frac{1}{λ}$ for light $]$

$∴$ $\frac{λ_{N a}}{λ_{C u}} = \frac{E_{C u}}{E_{N a}} = \frac{4.5}{2.3} \approx \frac{2}{1}$