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Radius of certain orbit in hydrogen atom is $4.76 A^{o}$ , total energy of electron revolving in that orbit will be

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$- 2.42 \times 10^{- 19} J$

$- 7.29 \times 10^{- 19} J$

$- 6.54 \times 10^{- 18} J$

$- 1.36 \times 10^{- 19} J$

answer is A.

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Detailed Solution

$r = 0.529 \times \frac{n^{2}}{Z} \Rightarrow 4.76 = 0.529 \times n^{2}$

$n = \sqrt{\frac{4.76}{0.529}} = \sqrt{9} = 3$

$E = - 2.18 \times 10^{- 18} \times \frac{Z^{2}}{n^{2}} = - 2.18 \times 10^{- 18} \times \frac{1}{9}$

$= - 2.42 \times 10^{- 19}$

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