First slide

Photo Electric Effect

Question

When photons of energy hv fall on an aluminium plate (of work function $E_{o}$ ), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

Moderate

A

K + $E_{o}$
B

2K
C

K
D

K + h $ν$

Solution

According to Einstein,s photoelectric effect energy of photon : KE of photoelectron + word function of metal. That is,
$\begin{array}{l} hc = \frac{1}{2} {mv}^{2} + E_{o} or hc = K_{\max} + E_{o} . . . (i) \\ Now, we have given, v' = 2 v \\ Therefore, K_{\max}^{'} = h (2 v) - E_{o} \\ K_{\max}^{'} = 2 hv - E_{o} . . . (ii) \\ From Eq . (i) and (ii), we have K_{\max}^{'} = 2 (K_{\max} + E_{o}) - E_{o} = 2 K_{\max} + E_{o} = K_{\max} + (K_{\max} + E_{o}) \\ = K_{\max} + hc [from Eq . (i)] \\ Putting K_{\max} = K \Rightarrow K_{\max}^{'} = K + hv \end{array}$

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