Answer is (D)
Using $\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}$
When, wavelength $\lambda$ is incident on metallic surface, the stopping potential required to stop most energetic electron is $\mathrm{V}$ and when wavelength increases to $3 \lambda$, stopping potential required as $\frac{\mathrm{V}}{6}$ $\therefore \frac{\mathrm{hc}}{\lambda}=\mathrm{eV}+\phi \quad \ldots(1)$
And $\frac{\mathrm{hc}}{3 \lambda}=\frac{\mathrm{eV}}{6}+\phi \quad \ldots(2)$
Thus equation (2) can be rewritten as $\frac{2 \mathrm{hc}}{\lambda}=\mathrm{eV}+6 \phi$
Subtracting equation (1) from (3) we get
$\frac{\mathrm{hc}}{\lambda}=5 \phi$ and $\phi=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$
Thus $\lambda_{0}=5 \lambda$
When light of wavelength ‘ $\lambda$ ‘ is incident on photosensitive surface, the stopping potential is ‘ $\mathrm{V}$ ‘. When light of wavelength ‘ $3 \lambda$ ‘ is incident on same surface, the stopping potential is $\frac{\mathrm{V}^{\prime}}{6}$. Threshold wavelength for the surface is
A) $2 \lambda$
B) $3 \lambda$
C) $4 \lambda$
D) $5 \lambda$
A) $2 \lambda$
B) $3 \lambda$
C) $4 \lambda$
D) $5 \lambda$
When light of wavelength ‘ $\lambda$ ‘ is incident on photosensitive surface, the stopping potential is ‘ $\mathrm{V}$ ‘. When light of wavelength ‘ $3 \lambda$ ‘ is incident on same surface, the stopping potential is $\frac{\mathrm{V}^{\prime}}{6}$. Threshold wavelength for the surface is
A) $2 \lambda$
B) $3 \lambda$
C) $4 \lambda$
D) $5 \lambda$
A) $2 \lambda$
B) $3 \lambda$
C) $4 \lambda$
D) $5 \lambda$