For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3V_0$. When the same metal is illuminated by light of wavelength $2\lambda$, then the stopping potential becomes $V_0$. The threshold wavelength for photoelectric emission for the given metal is $\alpha\lambda$. The value of $\alpha$ is ______.
Answer: (B) $4$
Einstein's equation for the two cases:
$\dfrac{hc}{\lambda}=\phi+3eV_0,\qquad\dfrac{hc}{2\lambda}=\phi+eV_0$
Subtracting: $\dfrac{hc}{2\lambda}=2eV_0\Rightarrow eV_0=\dfrac{hc}{4\lambda}$.
Then $\phi=\dfrac{hc}{2\lambda}-\dfrac{hc}{4\lambda}=\dfrac{hc}{4\lambda}$, so the threshold wavelength is $\lambda_0=\dfrac{hc}{\phi}=4\lambda$ and $\alpha=4$.
Solution by Sreeraj P, M.Sc Physics