JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Modern Physics

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Å is 0.42 V. If the threshold frequency is

x \times 1 0^{13}

^{- 1}

, where x is ______ (nearest integer).
(Given, speed light

= 3 \times 1 0^{8}

m/s, Planck's constant

= 6.63 \times 1 0^{- 34}

Js)

Answer: 35

Solution

From Einstein's photoelectric equation:

E = ϕ + e V_{0}

where

E

is the incident photon energy,

ϕ

is the work function, and

V_{0} = 0.42

V is the stopping potential.

The incident photon energy:

E = \frac{h c}{λ} = \frac{12400 eV\cdot A ˚}{6630 A ˚} \approx 1.87 eV

Work function:

ϕ = E - e V_{0} = 1.87 - 0.42 = 1.45 eV

Threshold frequency:

ϕ = h ν_{0}

1.45 \times 1.6 \times 1 0^{- 19} = 6.63 \times 1 0^{- 34} \times ν_{0}

ν_{0} = \frac{2.32 \times 1 0 ^{- 19}}{6.63 \times 1 0 ^{- 34}} \approx 3.5 \times 1 0^{14} = 35 \times 1 0^{13} s^{- 1}

Therefore,

x = 35

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer