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Two identical metal plates show photoele...

Two identical metal plates show photoelectric effect. Light of wavelength `lamda_A` falls on plate A and `lamda_B` fall on plate B and `lamda_A=2lamda_B`, The maximum KE of the photoelectrons are `K_A` and `K_B`, respectively, Which one of the following is true?

A

`2K_A=K_B`

B

`K_A lt K_B//2`

C

`K_A =2K_B`

D

`K_Agt K_B//2`

Text Solution

Verified by Experts

The correct Answer is:
B
`K_A =(hc)/lamda_A - phi_0 = (hc)/(2lamda_B) -phi_0`
`K_B = (hc)/lamda_B-phi_0`
Now, `K_A =1/2 ((hc)/lamda_B -phi_0) -phi_0 +phi_0/2`
`K_A =1/2 K_B -1/2 phi_0`
`K_B/2 gt K_A implies K_B gt 2K_A`
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