The kinetic energy of the photoelectrons increases by 0.52 eV when the wavelength of incident light is changed from 500 nm to another wavelength which is approximately

Question

The kinetic energy of the photoelectrons increases by 0.52 eV when the wavelength of incident light is changed from 500 nm to another wavelength which is approximately (A) 1250 nm (B) 1000 nm (C) 700 nm (D) 400 nm

Tardigrade Admin · Accepted Answer

KE 1- KE 2=( hc /λ1)-( hc /λ2) Δ KE = hc [(1/λ1)-(1/λ2)]

Q. The kinetic energy of the photoelectrons increases by $0.52 e V$ when the wavelength of incident light is changed from $500 nm$ to another wavelength which is approximately

1250 nm

1000 nm

700 nm

400 nm

$K E_{1} - K E_{2} = \frac{h c}{λ _{1}} - \frac{h c}{λ _{2}}$
$Δ K E = h c [\frac{1}{λ _{1}} - \frac{1}{λ _{2}}]$

Q. The kinetic energy of the photoelectrons increases by 0.52eV when the wavelength of incident light is changed from 500nm to another wavelength which is approximately