The variation of stopping potential as a function of the frequency of the incident light for a metal is shown in figure. The work function of the surface is [2023]

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Solution

Q.
The variation of stopping potential $(V_{0})$ as a function of the frequency $(ν)$ of the incident light for a metal is shown in figure. The work function of the surface is [2023]

1 18.6 eV

2 2.98 eV

3 2.07 eV

4 1.36 eV

Ans.
(3)

$e V_{0} = h ν - ϕ$

$0 = h ν - ϕ$

$ϕ = h ν = 6.6 \times 10^{- 34} \times 5 \times 10^{14} = 33 \times 10^{- 20} J$

$ϕ = \frac{33 \times 10^{- 20}}{1.6 \times 10^{- 19}} = 2.07 eV$

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