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Byju's Answer

Standard XII

Physics

Cut-off Wavelength

When photons ...

Question

When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, $T_{A}$ expressed in eV and de-Broglie wavelength $λ_{A}$ . The maximum kinetic energy of photoelectrons liberated from another metal B by photons 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} = 2.0 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 $T_{A} = 2.0 e V$
From photoelectric equation for the two metals
$4.25 e V - ϕ_{A} = T_{A} = \frac{p_{A}^{2}}{2 m} = \frac{h^{2}}{2 m λ_{A}^{2}}$
and $4.70 e V - ϕ_{B} = T_{B} = \frac{h^{2}}{2 m λ_{B}^{2}}$
$\frac{T_{A}}{T_{B}} = {(\frac{λ_{B}}{λ_{A}})}^{2} = 4$ , given $T_{B} = T_{A} - 1.50 e V$
On solving we get, $T_{A} = 2.0 e V$ and $T_{B} = 0.50 e V$
$∴ ϕ_{A} = 2.25 e V, ϕ_{B} = 4.20 e V$

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