The wave motion of electron in a Bohr's orbit of hydrogen is as shown in diagram. The potential energy of electron in the given orbit of hydrogen atom is:

- 3.4

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+ 3.4

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- 3.02

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- 1.51

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Solution

The correct option is C $- 3.02$ eV
$n = 3$
$P E = - 27.2 \times \frac{z^{2}}{n^{2}} e V$
$P E = - 27.2 \times \frac{1^{2}}{3^{2}} e v$
$P E = - 3.02 e V$

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The wave motion of electron in a Bohr's orbit of hydrogen is as shown in diagram. The potential energy of electron in the given orbit of hydrogen atom is:

The correct option is C −3.02 eVn=3PE=−27.2×z2n2eVPE=−27.2×1232evPE=−3.02eV

The correct option is C $- 3.02$ eV
$n = 3$
$P E = - 27.2 \times \frac{z^{2}}{n^{2}} e V$
$P E = - 27.2 \times \frac{1^{2}}{3^{2}} e v$
$P E = - 3.02 e V$