Question

When a certain metal was irradiated with light of frequency 1.6*10^16 hz?

Answer 1

Answer:

KE1=h(V1−V0)------(i)

KE2=h(V2−V0)=KE12------(i)

Dividing equations (ii) by (i) we have

∴V2−V0V1−V0=12

1.0×1016−V01.6×1016−V0=12

2.0×1016−2V0=1.6×1016−V0

V0=4×1015Hz

Hence (c) is the correct answer.

Explanation:

Answer 2

Answer:

for frequency= 1.6*10^16s-¹

E1= h υ

= h*1.6*10^16s-¹

KE= hυ+hυo

KE1=h*1.6*10^16s-¹+hυo

for frequency= 1.0"10^16s-¹

E2= h*1.0*10^16s-¹

KE2=h*1.0*10^16s-¹+hυo

according to question

KE1=2KE2

h*1.6*10^16s-¹+hυo=2(h*1.0*10^16s-¹+hυo)

h*1.6*10^16s-¹+hυo=2*h*1.0*10^16s-¹+2hυo

h(1.6-1.0)*10^16=hυo

υo=0.6*10^16s-¹

= 6*10^15s-¹

Explanation: