Question

When a metal is irradiated with light of
frequency 4.0x10^16 second inverse the photo electrons emitted had six times the K.E as the K.E of photo electron emitted when the metal was
irradiated with light of frequency 2.0 10^16 second inverse The calculate the critical frequency of the
metal.
1)2.0x10^16 2)1.6x10^16
3)3.0x10^16 4)4.2x10^16​

Answer 1

Answer:

option c is the right one

Explanation:

no need 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

,,,,,,,,ooo,,,,,oooo,,,oooooooo,,,,

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: