MCQ
What is the potential energy of the electron in the $L-$ shell of the hydrogen atom? .............. $\mathrm{eV}$
  • A
    $-13.6$
  • $-6.8$
  • C
    $-10.2$
  • D
    $-3.4$

Answer

Correct option: B.
$-6.8$
b
The principal quantum number for L-shell is 2 we know that energy of an electron in $\mathrm{nth}$ bohr's orbit is given by, $E_{n}=-\frac{13.6 \mathrm{eV}}{n^{2}}$

but Potential energy of electron $=-2\left(E_{n}\right)=-\frac{27.2 \mathrm{eV}}{n^{2}}$ Therefore, for $n=2$

$P.E. =-\frac{27.2 \mathrm{eV}}{4}=-6.8 \mathrm{eV}$

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