Physics - Modern Physics Question with Solution | TestHub

Photoemission will stop when potential on silver sphere becomes equal to the stopping potential. $\therefore \frac{hc}{\lambda }-W=e{V}_0$ Here, $V_0=\frac{1}{4\pi {\epsilon }_0}\frac{ne}{r}$ $$\begin{array}{rl}& \therefore \left(\frac{1240}{1200}\mathrm{eV}\right)-(4.7\mathrm{eV})\\ & =\frac{9\times 10^9\times n\times 1.6\times 10^{-19}}{10^{-2}}\\ & (6.2-4.7)=\frac{9\times 10^9\times n\times 1.6\times 10^{-19}}{10^{-2}}\\ & \text{ or }n=\frac{1.5\times 10^{-2}}{9\times 1.6\times 10^{-10}}=1.04\times 10^7\end{array}$$ $\therefore$ Answer is 7 . Analysis of Question (i) Question is moderately difficult. (ii) In ${\mathrm{eV}}_0$ if we substitute the value of $e=1.6\times 10^{-19}\mathrm{C}$, then answer is in J. If we leave it then, answer is in $e\mathrm{V}$.