Question

Wavelength λA, and λB are incident on two identical metal plates and photo electrons are emitted. If λA = 2λB , the maximum kinetic energy of photo electrons is _______ . (A) 2KA = KB (B) KA < KB/2 (C) KA = 2KB (D) KA > 2KB /2

Answer 1

Answer:

Option C is correct ........

Answer 2

Answer:

$(A)\ 2K_A\ =\ K_B$

Explanation:

Let's the first wavelength be $\lambda_A$ and the second wavelength be $\lambda_B.$

According to the theory of photoelectric effect the maximum kinetic energy of photo electrons can be given by

$K\ =\ \dfrac{hc}{\lambda}$

where, h = Planck's constant

            c = speed of light

From the question,

$\lambda_A\ =\ 2\lambda_B$

Hence, the maximum kinetic energy for first wavelength,

$K_A\ =\ \dfrac{h.c}{\lambda_A}$

$=>\ K_A\ =\ \dfrac{h.c}{2.\lambda_B}$

$=>\ K_A\ =\ \dfrac{K_B}{2}$

$=>\ 2.K_A\ =\ K_B$

Hence, the maximum kinetic energy of photo electrons is

$2.K_A\ =\ K_B$

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