Question

The radius of electron orbit and the speed of electron in the ground state of hydrogen atom is $5.30\times 10^{-11}m$ and $2.2\times 10^{6}m{s}^{-1}$ respectively, then the orbital period of this electron in second excited state will be

• A. $1.21\times 10^{-14}s$
• B. $1.21\times 10^{-12}s$
• C. $1.21\times 10^{-10}s$
• D. $1.21\times 10^{-15}s$

Answer 1

Answer: (D)

Here, $r_1=5.30\times 10^{-11}m$;
$v_1=2.2\times 10^{6}m{s}^{-1}$
In the second excited state,
$r_n=n^2{r}_1,v_n=\frac{v_1}{n}$
$\therefore r_2=4r_1=4\times 5.30\times 10^{-11}m$
$=2.12\times 10^{-10}m$
ans $v_2=\frac{v_1}{2}$
$=\frac{2.2\times 10^6}{2}m{s}^{-1}$
$=1.1\times 10^{6}m{s}^{-1}$
Since, orbit period
$(T)=\frac{2\pi r_2}{v_2}$
$=\frac{2\times 3.14\times 2.12\times 10^{-10}}{1.1\times 10^6}$
$=\frac{13.31\times 10^{-16}}{1.1}$
$=1.21\times 10^{-15}s$