When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 𝑉. This potential drops to 0.6 𝑉 if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take $\frac{hc}{e} = 1 .24 \times 10^{- 6} {JmC}^{- 1} .$ ]

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$1 .72 \times 10^{- 7} m, 1 .20 eV$

$3 .78 \times 10^{- 7} m, 5 .60 eV$

$1 .72 \times 10^{- 7} m, 5 .60 eV$

$3 .78 \times 10^{- 7} m, 1 .20 eV$

answer is A.

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Detailed Solution

$\begin{aligned} h v - ϕ = 6 eV \\ \frac{h c}{λ} - ϕ = 6 eV \\ \frac{h c}{4 λ} - ϕ = 0.6 eV \\ \frac{3 h c}{4 λ} = 5.4 eV \end{aligned}$
$\begin{array}{l} ∴ λ = \frac{3 h c}{4 \times 5.4 eV} = \frac{3 \times 1.24 \times 10^{- 6}}{4 \times 5.4} \\ = 1.72 \times 10^{- 7} m \\ \Rightarrow from equation (i) \\ \frac{h c}{1.72 \times 10^{- 7}} \times \frac{1}{1.6 \times 10^{- 19}} - ϕ = 6 eV \\ \frac{2 \times 10^{- 25}}{2.75 \times 10^{- 26}} - ϕ = 6 \\ \Rightarrow ϕ = (7.27 - 6) ≅ 1.2 eV \end{array}$