Question

As the electron in Bohr’s orbit of hydrogen atom passes from state $n$ = 2 to, $n =1$, the Kinetic energy (K) and the potential energy (U) changes as

• A. $K$ four fold, $U$ also four fold
• B. $K$ two fold, $V$ also two fold
• C. $K$ four fold, $U$ two fold
• D. $K$ two fold, $U$ four fold

Answer 1

Answer: (A)

KE of an electron in $n$th orbit : $K_{n} \propto \frac{1}{n^{2}}$ and
PE of an electron in $n$th orbit : $U_{n} \propto \frac{1}{n^{2}}$
$\therefore $ When an electron passes from state $n=2$ to $n=1$
$\frac{K_{2}}{K_{1}} =\frac{1^{2}}{2^{2}}=\frac{1}{4} $
$K_{1} =4 K_{2}$
$\frac{U_{2}}{U_{1}}=\frac{1^{2}}{2^{2}}=\frac{1}{4}$
or $ U_{1}=4 U_{2}$