Question
For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3{\text{ }V}_{0}$. 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 $\_\_\_\_$ .
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Solution
Sol. [Key Concept: Use $\frac{hc}{\lambda} - \varphi = eV$ for both wavelengths; solve for $\varphi$ and $\lambda_{0}$.]
$\frac{hc}{\lambda} - \phi = 3{eV}_{0}\ldots\ldots\ldots$ (i)
$\frac{hc}{2\lambda} - \phi = {eV}_{0}\ldots\ldots..($ ii $)$
$$\frac{hc}{\lambda} - \phi = 3\left\lbrack \frac{hc}{2\lambda} - \phi \right\rbrack $$
$$\phi = \frac{hc}{4\lambda} $$
$$\frac{hc}{\lambda_{0}} = \frac{hc}{4\lambda} $$
$$\lambda_{0} = 4\lambda $$
$$\alpha = 4 $$
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