Question

If speed of electron in ground state energy level is $2.2 \times 10^6\, m\, s^{-1}$ then its speed in fourth excited state will be

• A. $6.8 \times 10^6 \,m\,s^{-1}$
• B. $8.8 \times 10^5 \,m\,s^{-1}$
• C. $5.5 \times 10^5 \,m\,s^{-1}$
• D. $5.5 \times 10^6 \,m\,s^{-1}$

Answer 1

Answer: (C)

According to Bohr’s model
$v= \frac{2Ke^{2}Z}{nh} $
or $ v \propto \frac{1}{n} $
$ \therefore \frac{v_{A}}{V_{B}} = \frac{n_{B}}{n_{A}} $
Here, $v_{A} = 2.2 \times10^{6} ms^{-1}; $
$ n_{A} = 1, n_{B} = 4$
$ \therefore v_{B} = v_{A} \times \frac{n_{A}}{n_{B}} $
$= 2.2 \times10^{6} \times\frac{1}{4}$
$= 0.55 \times10^{6} $
$= 5.5 \times 10^{5} ms^{-1} $