The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491nm is 0 . 710V . When the incident wavelength is changed to a new value, the stopping potential is 1 . 43V

Correct answer: A. 382nm (hc)/(λ) = ϕ + e V _ s (1240)/(491) = ϕ + 0 . 71 . . . (1) (1240)/(λ) = ϕ + 1 . 43 . . . (2) ∴ λ = 382nm

JEE Main 202125 FebruaryEvening Shift

JEE Main 2021 · 25 February · Evening Shift · Physics · Dual Nature of Matter

The stopping potential for electrons emitted from a photosensitive...

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength

491nm

0 . 710V .

When the incident wavelength is changed to a new value, the stopping potential is

1 . 43V .

The new wavelength is :

A. $382nm$

$\dfrac{hc}{\lambda} = ϕ + e{V}_{s}$

$\dfrac{1240}{491} = ϕ + 0 . 71 . . . \left(1\right)$

$\dfrac{1240}{\lambda} = ϕ + 1 . 43 . . . \left(2\right)$

$∴ \lambda = 382nm$