Question

The radius of orbit of an electron and the speed of electron in the ground state of hydrogen atom are $5.5\times 10^{-11}m$ and $4\times 10^{6}m{s}^{-1}$ respectively. Then, the orbital period of this electron in the first excited state will he $\dots$

• A. $6.908\times 10^{-16}s$
• B. $9.608\times 10^{-16}s$
• C. $7.806\times 10^{-16}s$
• D. $8.9068\times 10^{-16}s$

Answer 1

Answer: (A)

Given, in ground state the radius and velocity of electron in hydrogen atom
$r_1=5.5\times 10^{-11}m$
$V_1=4\times 10^{6}m/s$
In first excited state
$r_n\propto \frac{n^2}{Z}$
$r_n=5.5\times 10^{-11}\times 2^2$
$=22.0\times 10^{-11}m/s$
The speed of electron
$v_n\propto \frac{Z}{n}$
$=\frac{4\times 10^6}{2}$
$=2\times 10^{6}m/s$
The time period of revolution
$T=\frac{2\pi r}{v_n}$
$=\frac{2\times 3.14\times 22.0\times 10^{-11}}{2\times 10^6}$
$=6.908\times 10^{-16}s$