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,
$r_n = n^2 r_1 , v_n = \frac{v_1}{n}$
$\therefore r_2 = 4 r_1 = 4\times 5.30 \times 10^{-11} m$
$= 2.12 \times 10^{-10} m$
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$