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An electron in the hydrogen atom jumps from excited state $n$ to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function $2.75 \mathrm{eV}$. If the stopping potential of the photo-electron is $10 \mathrm{~V}$, the value of $n$ is
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The correct answer is:
3
$$
\begin{aligned}
E & =\mathrm{KE}_{\max }+W \\
& =e V_0+W \\
& =10+2.75 \\
E & =12.75 \mathrm{eV}
\end{aligned}
$$
Difference of 4 and 1 energy level is $12.75 \mathrm{eV}$. So, higher energy level is 4 to ground and excited state is $n=3$.
\begin{aligned}
E & =\mathrm{KE}_{\max }+W \\
& =e V_0+W \\
& =10+2.75 \\
E & =12.75 \mathrm{eV}
\end{aligned}
$$
Difference of 4 and 1 energy level is $12.75 \mathrm{eV}$. So, higher energy level is 4 to ground and excited state is $n=3$.
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