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S1 : When ultraviolet light is incident on a photocell, its stopping potential is $V_{\circ}$ and the maximum kinetic energy of the photoelectrons is $K_{max}$ increase. . When the ultraviolet light is replaced by x-rays, both $V_{\circ}$ and $ K_{max}$ S2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

$\begin{array}{1 1} \text{(a) S1 is true, S2 is true; S2 is correct explanation of S1.}\\ \text{(b) S1 is true, S2 is true; S2 is not correct explanation of S1.} \\ \text{(c)S1 is false; S2 is true.}\\ \text{(d) S1 is true; S2 is false.}\end{array}$
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Ans : (d)
Since the frequency of ultraviolet light is less than the frequency of X-rays, the
energy of each incident photon will be more for X-rays
$KE_{photoelectron} = hv – \phi$
Stopping potential is to stop the fastest photoelectron
$V_{\circ} = \large\frac{hv}{e} – \large\frac{\phi}{e}$
and $V_{\circ}$ both increases. So, $KE_{max}$
But $KE$ ranges from zero to $KE_{max}$ because of loss of energy
due to subsequent collisions before getting ejected and not due to
range of frequencies in the incident light.


answered Dec 30, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1

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