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If p is the momentum of the fastest elec...

If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength `lambda` then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function):

A

`3/4 lamda`

B

`1/2 lamda`

C

`4/9 lamda`

D

`2/3 lamda`

Text Solution

Verified by Experts

The correct Answer is:
C
As kinetic energy is much higher than work function so `E= E_0 + KE ~~ KE`
As `E =(hc)/lamda` and KE `=(P^2/(2m))`
So, `lamda_2/lamda_1=(P_1/P_2)^2 implies lamda_2/lamda=(1/(1.5))^2`
`lamda_2=4/9lamda`
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