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
$v=\frac{2K{e}^{2}Z}{nh}$
or $v\propto \frac{1}{n}$
$\therefore \frac{v_A}{V_B}=\frac{n_B}{n_A}$
Here, $v_A=2.2\times 10^{6}m{s}^{-1};$
$n_A=1,n_B=4$
$\therefore v_B=v_A\times \frac{n_A}{n_B}$
$=2.2\times 10^6\times \frac{1}{4}$
$=0.55\times 10^6$
$=5.5\times 10^{5}m{s}^{-1}$