Q82 / 120 · NEET 2010 (Mains) · Physics

The electron in the hydrogen atom jumps from excited state $(n=3)$ to its ground state $(n=1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \mathrm{eV}$, the stopping potential is estimated to be (the energy of the electron in $n$th state $\mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^2} \mathrm{eV}$ )

  • A
    $5.1 \mathrm{~V}$
  • B
    $12.1 \mathrm{~V}$
  • C
    $17.2 \mathrm{~V}$
  • Correct answer
    D
    $7 \mathrm{~V}$

Solution

For $n=1, \mathrm{E}_1=-\frac{13.6}{(1)^2}=-13.6 \mathrm{eV}$ and for $n=3, E_3=-\frac{13.6}{(3)^2}=-1.51 \mathrm{eV}$ So, required energy $\begin{array}{l} \mathrm{E}=\mathrm{E}_3-\mathrm{E}_1 =-(1.51)-(-13.6) =12.09 \mathrm{eV} \\ \because \mathrm{E} =\mathrm{W}+\mathrm{eV} \\ \therefore \mathrm{eV} =\mathrm{E}-\mathrm{W} \\ \mathrm{eV} =(12.09-51) \mathrm{e} =7 \text { volt } \end{array}$

Hence, the correct answer is option: 'D'.

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