1.

Two identical metal plates show photoelectric effect. Light of wavelength `lamda_A` falls on plate A and `lamda_B` fall on plate B `lama_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=2K_B`C. `(K_AltK_B/(2))`D. `K_Agt2K_B`

Answer» Correct Answer - C
`K_A=(hc)/(lamda_A)-phi_0`,`K_B=(hc)/(lamda_B)-phi_0`
But `lamda_A=2lamda_B`, therefore
`K_A=(hc)/(2lamda_B)-phi_0`
`K_A=(1)/(2)[K_B+phi_0]-phi_0`
or `K_A=(K_B)/(2)-(phi_0)/(2)`
`K_Alt(K_B)/(2)`


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