Question

Calculate the max.K.E of photoelectron emitted when a light of frequency 2×10^16 Hz is irradiated on a metal surface with threshold frequency equal to 8.68×10^15Hz??

Answer 1

Q)

 

The maximum kinetic energy of photoelectrons ejected from a metal,when it is irradiated with radiation of frequency 2×1014s−12×1014s−1 is 6.63×10−20J6.63×10−20J.The threshold frequency of metal is

(a)2×1014s−1(c)1×10−14s−1(b)3×1014s−1(d)1×1014s−1(a)2×1014s−1(b)3×1014s−1(c)1×10−14s−1(d)1×1014s−1



A)

 



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

Answer:

E(Total Energy)=Eo(Minimum Energy reqd. for the electron)+K.E(kinetic Energy of Photo electron)

hv=hv0 + 1/2 mv2

6.6*10-34*(3.2 * 1016) =Eo+ K.E

 2.112*10-17 =Eo+ K.E......................(1)

6.6*10-34*(2*1016) =Eo+ 2*K.E.

1.32*10-17=Eo+ 2 K.E  .......................(2)

solving equation (1) and (2):

Eo=2.904*10-17 j

Explanation: