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 $4 \times 10^{6} ms ^{-1}$ respectively. Then, the orbital period of this electron in the first excited state will he $\ldots$

• 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
$T =\frac{2 \pi r}{v_{n}}$
$=\frac{2 \times 3.14 \times 22.0 \times 10^{-11}}{2 \times 10^{6}}$
$=6.908 \times 10^{-16} s$