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Byju's Answer

Standard XII

Physics

Cut-off Wavelength

When photon o...

Question

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

The work function of A is 2.25 eV

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The work function of B is 4.20 eV

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T_{A} = 2.00 e V

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T_{B} = 2.75 e V

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Solution

The correct options are
A The work function of A is 2.25 eV
B The work function of B is 4.20 eV
C $T_{A} = 2.00 e V$
$K_{m a x} = E - W_{0}$
$∴ T_{A} = 4.25 - (W_{0})_{A}$ ...(i)
$T_{B} = (T_{A} - 1.5) = 4.70 - (W_{0})_{B}$ ...(ii)
Equation (i) and (ii) gives $(W_{0})_{B} - (W_{0})_{A} = 1.95 e V$
De Broglie wave length
$λ = \frac{h}{\sqrt{2 m k}} \Rightarrow λ \propto \frac{1}{\sqrt{K}}$
$\Rightarrow \frac{λ_{B}}{λ_{A}} = \sqrt{\frac{K_{A}}{K_{B}}} \Rightarrow 2 = \sqrt{\frac{T_{A}}{T_{A} - 1.5}} \Rightarrow T_{A} = 2 e V$
From equation (i) and (iii)
$W_{A} = 2.25 e V$ and $W_{B} = 4.20 e V$ .

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