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Q. In the hydrogen atom, the electron is moving around the nucleus with velocity $2.18 \, \times \, 10^{6} \, m \, s^{- 1}$ in an orbit of radius $0.528 \, \overset{^\circ }{A}$ . The acceleration of the electron is

NTA AbhyasNTA Abhyas 2022

Solution:

Given, $v=2.18 \, \times 10^{6} \, m \, s^{- 1}, \, r=0.528 \, \times \, 10^{- 10 \, }m$
Acceleration of electron moving round the nucleus
$ \, a \, =$ $\frac{\left(2.18 \, \times \left(10\right)^{6}\right)^{2}}{0.528 \, \times \left(10\right)^{- 10}}\approx$ $9 \, \times 10^{22} \, m \, s^{- 2}$