Question

Derive an expression of the energy of the electron in the nth orbit of hydrogen atom

Answer 1

Explanation:

1. this is the explanation of the answer

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Answer 2

SOLUTION

$e \: = \: \frac{2\pi {}^{2} \: z {}^{2}e {}^{4}m }{n {}^{2} {h}^{2} }$

Z = atomic number , e = charge of electron

m = mass of electron ,n = principal of quantum

h = Planck constant

$e \: = - \frac{13.6 \: \times {z}^{2} }{ {n}^{2} } \: ev \:$

for hydrogen atom Z = 1

$e \: = \frac{1.36}{ {n}^{2} } ev$