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When photons of energy $4.25 e V$ strike the surface of metal $A$ , the ejected photoelectrons have maximum kinetic energy, $T_{A}$ expressed in $e V$ and de-Broglie wavelength $λ_{A}$ . The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photons of energy $4.70 e V$ is $T_{B} = (T_{A} - 1.50 e V)$ . If de-Broglie wavelength of these photoelectrons is $λ_{B} = 2 λ_{A}$ , then find
the work function $W_{A}$ of metal $A$ and the work function $W_{B}$ of metal $B$ .

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$2.25 e v, 0.5 e v$

$2.25 e v 2.5 e v$

1.25 ev, $5 e v$

$0.5 e v, 1 e v$

answer is A.

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Detailed Solution

$K_{\max} = E - W$

Therefore, $T_{A} = 4.25 - W_{A}$ , $T_{B} = (T_{A} - 1.50) = 4.70 - W_{B}$

equations (1) and (2) gives,

$W_{B} - W_{A} = 1.95 e V$

de Broglie wavelength is given by

$λ = \frac{h}{\sqrt{2 K m}}$

$λ \propto \frac{1}{\sqrt{K}}$

$K = K E of electron$

$\Rightarrow \frac{λ_{B}}{λ_{A}} = \sqrt{\frac{K_{A}}{K_{B}}}$

$\Rightarrow 2 = \sqrt{\frac{T_{A}}{T_{A} - 1.5}}$

$\Rightarrow T_{A} = 2 e V$

From equation (1), we get

$W_{A} = 4.25 - T_{A} = 2.25 e V$

From equation (3), we get

$W_{B} = W_{A} + 1.95 e V = (2.25 + 1.95)$

$\Rightarrow W_{B} = 4.20 e V$

$\Rightarrow T_{B} = 4.70 - W_{B} = 4.70 - 4.20 = 0.50 e V$

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