Question

Find the frequency of revolution of the electron in the first stationary orbit of H-atom

• A. $8\times {10}^{14}Hz$
• B. $8.6\times {10}^{10}Hz$
• C. $8.6\times {10}^{-10}Hz$
• D. $8.12\times {10}^{146Hz}$

Answer 1

The correct option is D $8.12\times {10}^{146Hz}$

$frame=\frac{1}{Timeperiod}$

Time perid$=\frac{Totaldistancecoverd}{velocity}=\frac{2\pi r}{v}$

frequency$=\frac{v}{2\pi r}$

The velocity $v_2$ and radius $r_2$ of the $2^{nd}$ both orbit

$r_n=(0.53\times {10}^{-10}{n}^2/2)m$

$r_2=0.53\times {10}^{-10}\left(2^2\right)/m=2.12\times {10}^{-100}m$

$v_n=2.165\times {10}^6\left(2/n\right)m/s$

$v_2=2.165\times {10}^6\left(1/2\right)=1.082\times {10}^{6}m/s$

frequency$=\frac{v_2}{2\pi r_2}=\frac{1.082\times {10}^6}{2\left(x\right)(2.12\times {10}^{-10})}$

$F=8013\times {10}^{16}Hz$