When 1 cm thick surface is illuminated with light of wavelength λ, the stopping potential is V. When the same surface is illuminated by light of wavelength 2λ, the stopping potential is V/3. Threshold wavelength for metallic surface is:
Answer & explanation
Correct answer: option 2
According to the question,
$eV=hc(\frac{1}{λ}-\frac{1}{λ_0})$ (i)
and $\frac{eV}{3}=hc(\frac{1}{2λ}-\frac{1}{λ_0})$ (ii)
Dividing eq. (i) by eq. (ii), we get
$3=\frac{(\frac{1}{λ}-\frac{1}{λ_0})}{(\frac{1}{2λ}-\frac{1}{λ_0})}$
or $3(\frac{1}{2λ}-\frac{1}{λ_0})=\frac{1}{2λ}-\frac{1}{λ_0}$
or $\frac{3}{2λ}-\frac{1}{λ}=\frac{3}{λ_0}-\frac{1}{λ_0}$
or $\frac{1}{2λ}=\frac{2}{λ_0}$
Threshold wavelength for metallic surface
$λ_0=4λ$