Dual Nature Of Radiation And Matter | Physics JEE previous year questions with Solutions

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Topic Question Set

Q 31 :

A metallic surface is illuminated with radiation of wavelength $λ$ , the stopping potential is $V_{0}$ . If the same surface is illuminated with radiation of wavelength $2 λ$ , the stopping potential becomes $\frac{V_{0}}{4}$ . The threshold wavelength for this metallic surface will be [2023]

$\frac{λ}{4}$
$4 λ$
$\frac{3}{2} λ$
$3 λ$

1

2

3

4

(4)

$From the equation of photoelectric effect$

$e V_{0} = \frac{h c}{λ} - ϕ_{0} = \frac{h c}{λ} - \frac{h c}{λ_{0}}$

$\frac{e V_{0}}{4} = \frac{h c}{2 λ} - \frac{h c}{λ_{0}}$

$\Rightarrow \frac{1}{4} (\frac{h c}{λ} - \frac{h c}{λ_{0}}) = \frac{h c}{2 λ} - \frac{h c}{λ_{0}}$

$\frac{1}{λ_{0}} - \frac{1}{4 λ_{0}} = \frac{1}{2 λ} - \frac{1}{4 λ}$

$\frac{3}{4 λ_{0}} = \frac{1}{4 λ} \Rightarrow λ_{0} = 3 λ$

Q 32 :

The difference between threshold wavelengths for two metal surfaces A and B having work function $ϕ_{A} = 9 eV$ and $ϕ_{B} = 4.5 eV$ in nm is: (Given, $h c = 1242 eV nm$ ) [2023]

264
138
276
540

1

2

3

4

(2)

$λ_{A} = (\frac{1242}{9}) = 138 nm$

$λ_{B} = (\frac{1242}{4.5}) = 276 nm$

$λ_{B} - λ_{A} = 138 nm$

Q 33 :

Given below are two statements: [2023]

Statement I: Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.

Statement II: Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.

In the light of above statements, choose the correct answer from the options given below:

Statement I is true but Statement II is false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Both Statement I and Statement II are false

1

2

3

4

(1)

UV rays have maximum frequency hence are most effective for emission of electrons from a metallic surface.

$K E_{\max} = h f - h f_{0}$

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