Question

when a metal is illuminated by wavelength

ƛ the stopping potential is 3v when wavelength 2

ƛ is incident on same surface stopping potential is V .threshold wavelength?

Answer 1

We know, $V_o=\frac{hc}{e}\left(\frac{1}{\lambda }\right)-\frac{\phi }{e}$
Where V_o is stoping potential.
$\phi$ is work function and lambda is the wavelength .
According to question,
$$\begin{gathered} 3V=\frac{hc}{e}\left(\frac{1}{\lambda }\right)-\frac{\phi }{e}.............\left(1\right) \\ and \\ V=\frac{hc}{e}\left(\frac{1}{2\lambda }\right)-\frac{\phi }{e}............\left(2\right) \end{gathered}$$
on eliminating V from both equation , we get
$\phi =\frac{1}{4}\frac{hc}{\lambda }$
Now we know , threshold wavelength is given by
${\lambda }_o=\frac{hc}{\phi }$
$$\begin{gathered} puttingthevalueof\phi intheaboveequationtofind{\lambda }_{o}weget \\ {\lambda }_o=4\lambda \end{gathered}$$
Hence threshold wavelength is 4 lambda.