The orbital period of the electron in Bohr’s hydrogen atom for \( n=1 \) orbit is
(Given: Bohr radius : \( 0.53\times10^{-10}\,\mathrm{m} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \sim 1.5\times10^{-16}\,\mathrm{s} \)
The orbital period is the circumference divided by the orbital speed, \( T = 2\pi r/v \). The speed in the first orbit follows from Bohr’s quantisation \( mvr = h/2\pi \) (n = 1), using the standard values \( h = 6.63\times10^{-34}\,\mathrm{J\,s} \) and \( m = 9.1\times10^{-31}\,\mathrm{kg} \).
\( \displaystyle v = \frac{h}{2\pi mr} = \frac{6.63\times10^{-34}}{2\times3.14\times9.1\times10^{-31}\times0.53\times10^{-10}} \)
\( v \approx 2.19\times10^{6}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle T = \frac{2\pi r}{v} = \frac{2\times3.14\times0.53\times10^{-10}}{2.19\times10^{6}} = \frac{3.33\times10^{-10}}{2.19\times10^{6}} \)
\( T \approx 1.5\times10^{-16}\,\mathrm{s} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and added the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
