Chemistry - Atomic Structure Question with Solution | TestHub

From graph, threshold wavelength $=240\mathrm{nm}$. H

ence work function $=\frac{4\times 10^{-15}\times 3\times 10^8}{240\times 10^{-9}}=5\mathrm{eV}$.

When incident wavelength is $2{\lambda }_0$.

$(\mathrm{KE}{)}_{\max }={\mathrm{E}}_{\mathrm{P}}-\varphi$

$10={\mathrm{E}}_{\mathrm{P}}-5$

${\mathrm{E}}_{\mathrm{P}}=15\mathrm{eV}$

Hence when incident wavelength is ${\lambda }_0$ then energy of photon $=30\mathrm{eV}$. $(\mathrm{KE}{)}_{\max }=30-5=25\mathrm{eV}$.

So stopping potential is 25 volt.