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The retarding potential necessary to stop the emission of photoelectrons, when a target material of work function 1.24e V is irradiated with light of wavelength $4.36 \times 10^{- 7} m$ is

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0.36 eV

4.08 eV

2.84 eV

1.60 eV

answer is D.

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Detailed Solution

$Φ = 1.24 e V, λ = 4.36 \times 10^{- 7} m$

$U = \frac{h c}{λ} - ϕ$

$= [\frac{(6.6 \times 10^{- 34}) (3 \times 10^{8})}{(4.36 \times 10^{- 7}) (1.6 \times 10^{- 19})} - 1.24] e V$

$U = \frac{6.6 \times 3}{4.36 \times 1.6} - 1.24 = \frac{19.8}{6.976} - 1.24$

$= [2.83 - 1.24] e V$

$U = 1.60 e V$

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