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

Question

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 h c e = 1 . 24 × 10 - 6 J mC - 1 .]

TestHub · Accepted Answer

According to the Einstein's equation of photoelectric effect for the first case,
$\frac{h c}{λ} - ϕ = 6 eV \dots (i)$
And for the second case,
$\frac{h c}{4 λ} - ϕ = 0.6 eV$
Therefore,
$\frac{3 h c}{4 λ} = 5.4 eV$
$\Rightarrow λ = \frac{3 h c}{4 \times 5.4 eV} = \frac{3 \times 1.24 \times 10^{- 6}}{4 \times 5.4}$
$\Rightarrow λ = 1.72 \times 10^{- 7} m$
$\Rightarrow$ from equation $(i)$
$\frac{h c}{1.72 \times 10^{- 7}} \times \frac{1}{1.6 \times 10^{- 19}} - ϕ = 6 eV$
$\frac{2 \times 10^{- 25}}{2.75 \times 10^{- 26}} - ϕ = 6$
$\Rightarrow ϕ = (7.27 - 6) ≅ 1.2 eV$

Physics - Modern Physics Question with Solution | TestHub

Options:

Doubts & Discussion