JEE Main 2025 Apr 7 Shift 2, Physics Q5: Introduction and Photoelectric Effect

JEE Main2025Apr 7, Shift 2Physics

A photo-emissive substance is illuminated with a radiation of wavelength $λ_{i}$ so that it releases electrons with de-Broglie wavelength $λ_{e}$ . The longest wavelength of radiation that can emit photoelectron is $λ_{0}$ . Expression for de-Broglie wavelength is given by :

(m : mass of the electron, h : Planck's constant and c : speed of light)

A
$λ_{e} = \frac{h}{2 m c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}$
B
$λ_{e} = \frac{h λ _{0}}{2 m c}$
C
$λ_{e} = \frac{h}{2 m c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}$
D
$λ_{e} = \frac{h λ _{i}}{2 m c}$

Solution

By Einstein's photoelectric equation, $K E = E_{p h o t o n} - W = \frac{h c}{λ _{i}} - \frac{h c}{λ _{0}}$ , where the work function equals $h c / λ_{0}$ (using the threshold wavelength).

Using $K E = \frac{p ^{2}}{2 m}$ and the de-Broglie relation $p = \frac{h}{λ _{e}}$ : $\frac{h ^{2}}{2 m λ _{e}^{2}} = h c (\frac{1}{λ _{i}} - \frac{1}{λ _{0}})$ .

Solving for $λ_{e}^{2}$ : $λ_{e}^{2} = \frac{h}{2 m c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}$ .

Taking the positive square root gives $λ_{e} = \frac{h}{2 m c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}$ .

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JEE Main 2025 Apr 7 Shift 2, Physics Q5: Introduction and Photoelectric Effect

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