When light of 3000 A0 is incident on sodium chloride, the stopping potential is 1.85 volt and when light of 4000 A0 is incident, then the stopping potential becomes 0.82 volt. The threshold wavelength (in A0 ) for sodium is:h=6.626×10-34 J-sec., e=1.6×10-19C

V=1e{hcλ−ϕ} ⇒hcλ=ϕ+eV⇒43=ϕ+1.85ϕ+0.82⇒ϕ=2.27 eV⇒λ0=6.626×10-34×3×1082.27×1.6×10-19=5473A0

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When light of 3000 $\overset{0}{A}$ is incident on sodium chloride, the stopping potential is 1.85 volt and when light of 4000 $\overset{0}{A}$ is incident, then the stopping potential becomes 0.82 volt. The threshold wavelength (in $\overset{0}{A}$ ) for sodium is:
$(h = 6.626 \times 10^{- 34} J - \sec ., e = 1.6 \times 10^{- 19} C)$

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answer is 5473.

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Detailed Solution

$\begin{array}{l} V = \frac{1}{e} {\frac{h c}{λ} - ϕ} ⇒ \frac{h c}{λ} = ϕ + e V \\ ⇒ \frac{4}{3} = \frac{ϕ + 1.85}{ϕ + 0.82} \\ \Rightarrow ϕ = 2.27 e V \\ \Rightarrow λ_{0} = \frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{2.27 \times 1.6 \times 10^{- 19}} = 5473 \overset{0}{A} \end{array}$