The angular momentum of an electron in the hydrogen atom is hπ, when h is planks constant. The potential energy of this electron is

L=nh2π n=2TE=−13.64=−3.4∴PE=2×TE=−6.8 eV

L=nh2π n=2TE=−13.64=−3.4∴PE=2×TE=−6.8 eV

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The angular momentum of an electron in the hydrogen atom is $\frac{h}{π}$ , when h is planks constant. The potential energy of this electron is

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-6.8 eV

-3.4 eV

-27.2 eV

-13.6 eV

answer is B.

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$L = n \frac{h}{2 π}$

$\begin{array}{l} n = 2 \\ T E = \frac{- 13.6}{4} = - 3.4 \\ ∴ P E = 2 \times T E = - 6.8 e V \end{array}$

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