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Byju's Answer

Standard XII

Physics

Energy Levels

When photons ...

Question

When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $(K E)_{A}$ and de-Broglie wavelength is $λ_{A}$ . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.7 eV is $(K E)_{B}$ , where $(K E)_{B} = (K E)_{A} - 1.5$ eV. If the de-Broglie wavelength of these photoelectrons is $λ_{B} (= 2 λ_{A})$ , then:

The work function of metal A is 2.25 eV

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The work function of metal B is 4.20 eV

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$(K E)_{A} = 2 e v$

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$(K E)_{B} = 2.75 e v$

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Solution

The correct options are
A
The work function of metal A is 2.25 eV

B
The work function of metal B is 4.20 eV

C
$(K E)_{A} = 2 e v$

$4.25 = (W_{0})_{A} + (K . E)_{A} 4.70 = (W_{0})_{B} + (K . E)_{A} - 1.5$
So, $(W_{0})_{B} - (W_{0})_{A} = 0.45 + 1.5 = 1.95 . . . (1)$
Now, $λ_{B} = 2 λ_{A}$
$\frac{h}{\sqrt{2 m (K . E .)_{B}}} = \frac{2 h}{\sqrt{2 m (K . E .)_{A}}}$
So, $(K . E .)_{A} = 4 (K . E .)_{B}$
$4.25 - (W_{0})_{A} = 4 [4.7 - (W_{0})_{B}]$
$4 (W_{0}) B - (W_{0})_{A} = 14.55 . . . . . . (2)$
On solving (1) and (2)
$(W_{0}) B = 4.2 e V$
$(W_{0}) A = 2.25 e V$
$(K . E .)_{A} = 2 e V$
$(K . E .)_{B} = 0.5 e V$

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