Question

The maximum kinetic energy of photoelectrons ejected from a metal, when it is irradiated with radiation of frequency 2×10¹⁴s^-1 is 6.63 ×10^-20J. what is the threshold frequency of the metal??

Answer 1

swer

Absorbed energy =Threshold energy + Kinetic energy of photoelectrons

hv=hv0+KEhv=hv0+KE

hv0=hv−KEhv0=hv−KE

6.626×10−34×v0=6.626×10−34×2×10−14−6.63×10−206.626×10−34×v0=6.626×10−34×2×10−14−6.63×10−20

V0=1.3252×10−19−6.63×10−206.626×10−34V0=1.3252×10−19−6.63×10−206.626×10−34

V0=9.99×1013V0=9.99×1013

⇒1014s−1⇒1014s−1

Hence (d) is the correct answer

Answer 2

Absorbed energy =Threshold energy + Kinetic energy

hv=hv0+KEhv=hv0+KE

hv0=hv−KEhv0=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^−20)÷(6.626×10^−34)

V0=9.99×10^13

⇒10^14s−1

hope it helps u