Q. When a metal surface is illuminated by a monochromatic light of wavelength λ, then the potential difference required to stop the ejection of electrons is 3V. When the same surface is illuminated by the light of wavelength 2λ , then the potential difference required to stop the ejection of electrons is V. Then for photoelectric effect, the threshold wavelength for the metal surface will be
(a) 6λ
(b) 4λ/3
(c) 4λ
(d) 8λ
Ans: c
Sol: $\large \frac{h c}{\lambda} = \phi + e (3 V)$ …(i)
$\large \frac{h c}{2 \lambda} = \phi + e (V)$ ….(ii)
Dividing (i) by (ii)
$\large 2 = \frac{\phi + 3 e V}{\phi + e V}$
2φ + 2 e V = φ + 3 e V
φ = e V
$\frac{h c}{\lambda_o} = e V $
$\large \lambda_o = \frac{h c}{e V}$ ….(iii)
Subtracting (i) and (ii)
$\large \frac{h c}{2 \lambda} = 2 e V $
$\large \frac{h c}{e V} = 4 \lambda $
From(iii) , λo = 4λ