Question

OJ
PROBLEM 21. The maximum kinetic energy of the
photoelectron emitted from a metal is 2.03 X 10 J when
light that has 656 nm wavelength shines on the surface.
Determine the threshold frequency vo for this metal.
[Ans. 1.51 x 1014 Hz]
Can anyine plzz help me with this..​

Answer 1

Explanation:

Absorbed energy =Threshold energy + Kinetic energy of photoelectrons

hv=hv0+KE

hv0=hv−KE

6.626×10−34×v0=6.626×10−34×2×10−14−6.63×10−20

V0=1.3252×10−19−6.63×10−206.626×10−34

V0=9.99×1013

⇒1014s−1