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A silver sphere of radius $1 \mathrm{~cm}$ and work function $4.7 \mathrm{eV}$ is suspended from an insulating thread in free-space. It is under continuous illumination of $200 \mathrm{~nm}$ wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^Z$ (where $1 < A < 10$ ). The value of $Z$ is
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Verified Answer
The correct answer is:
7
Photoemission will stop when potential on silver sphere becomes equal to the stopping potential.
$\therefore \quad \frac{h c}{\lambda}-W=e V_0$
Here, $\quad V_0=\frac{1}{4 \pi \varepsilon_0} \frac{n e}{r}$
$$
\begin{aligned}
& \therefore \quad\left(\frac{1240}{1200} \mathrm{eV}\right)-(4.7 \mathrm{eV}) \\
&=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\
&(6.2-4.7)=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\
& \text { or } \quad n= \frac{1.5 \times 10^{-2}}{9 \times 1.6 \times 10^{-10}}=1.04 \times 10^7
\end{aligned}
$$
$\therefore$ 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 $e \mathrm{~V}$.
$\therefore \quad \frac{h c}{\lambda}-W=e V_0$
Here, $\quad V_0=\frac{1}{4 \pi \varepsilon_0} \frac{n e}{r}$
$$
\begin{aligned}
& \therefore \quad\left(\frac{1240}{1200} \mathrm{eV}\right)-(4.7 \mathrm{eV}) \\
&=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\
&(6.2-4.7)=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\
& \text { or } \quad n= \frac{1.5 \times 10^{-2}}{9 \times 1.6 \times 10^{-10}}=1.04 \times 10^7
\end{aligned}
$$
$\therefore$ 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 $e \mathrm{~V}$.
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