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Q.

Potential energy of electron in fourth orbit of hydrogen atom is

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a

-1.70 eV

b

-1.51 eV

c

-3.02 eV

d

-0.85 eV

answer is B.

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Detailed Solution

$E_{n} = - \frac{Z^{2}}{n^{2}} E_{1}; Z = 1, n = 4 E_{4} = - \frac{E_{1}}{16} = - \frac{13.6}{16} ev = - 0.85 eV P . E = 2 T . E;$

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