Question

Energy of electron in the first Bohr orbit of $H$-atom is $-313.6$ $kcal$ ${mol}^{-1}$; then the energy in second Bohr orbit will be:

• A. $+313.6$ $kcal$ ${mol}^{-1}$
• B. $-78.4$ $kcal$ ${mol}^{-1}$
• C. $-34.84$ $kcal$ ${mol}^{-1}$
• D. $-12.5$ $kcal$ ${mol}^{-1}$

Answer 1

Answer: (B)

$-78.4$ $kcal$ ${mol}^{-1}$

The energy in second Bohr orbit will be $\frac{-313.6kcal/mol}{n^2}=\frac{-313.6kcal/mol}{2^2}=-78.4kcal/mol$