Question

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 $5.1eV$, the stopping potential is estimated to be (the energy of the electron in $n^{th}$ state $E_n=\frac{13.6}{n^2}eV)$

• A. $5.1V$
• B. $12.1V$
• C. $17.2V$
• D. $7V$

Answer 1

Answer: (D)

$7V$

Energy released when electron in the atom jumps from excited state ($n=3$) to ground state ($n=1$) is
$E=h\nu =E_3-E_1=\frac{-13.6}{3^2}-\left(\frac{-13.6}{1^2}\right)=\frac{-13.6}{9}+13.6=12.1$ eV

Therefore, stopping potential: $e{V}_0=h\nu -{\varphi }_0$$=12.1-5.1$ [$\because$ work function ${\varphi }_0=5.1$]

$\Rightarrow V_0=7V$