Question

when a photon of frequency 1.0 x 10^15 s-1 was allowed to hit a metal surface, an electron having 1.988× 10^-19 J
kinetic energy was emitted. Determine the kinetic energy of the electron ejected when a photon with
a wavelength equal to 600 nm hits the metal surface.​

Answer 1

from Einstein's photoelectric effect theory,

K = E - Φ

where , K is kinetic energy of ejected electron, E is energy of incident photon and Φ is work function.

given, K = 1.988 × 10^-19 J

frequency of incident photon, f = 1 × 10^15 s-¹

so, energy of incident photon, E = hf

= 6.626 × 10^-34 Js × 10^15 s-¹

= 6.626 × 10^-19 J

now, work function, Φ = E - K

= 6.626 × 10^-19 - 1.988 × 10^-19

= 4.638 × 10^-19 J

now, wavelength of incident photon , λ = 600nm = 6 × 10^-7 m

so, energy of incident photon, E' = hc/λ

= 6.626 × 10^-34 × 3 × 10^8/6 × 10^-7

= 3.313 × 10^-19 J

here we see, work function , Φ > energy of incident photon, E'

so, electron doesn't eject from metal surface.