Let be the orbital speed of the electron in a hydrogen atom in the ground-state level, =1.
For an electron, is given by the relation,
\(v_{1}=\frac{e^{2}}{n_{1} 4 \pi \epsilon_{0} \frac{h}{2 \pi}}=\frac{e^{2}}{2 \epsilon_{0} h}\)
where,
= Permittivity of free space =
h = Planck's constant =
\({v}_{1}=\frac{1.6 \times 10^{-19^{2}}}{2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}=0.0218 \times 10^{8}=2.18 \times 10^{6} \mathrm{~m} / \mathrm{s}\)
For level = 2, we can write the relation for the corresponding orbital speed as:
\(v_{2}=\frac{e^{2}}{n_{2}\times2 \epsilon_{0} h}=\frac{n_1}{2}=1.09 \times 10^{6} \mathrm{~m} / \mathrm{s}\)
And, for = 3, we can write the relation for the corresponding orbital speed as
\(v_{3}=\frac{e^{2}}{n_{3}\times2 \epsilon_{0} h}=\frac{n_1}{3}=7.27 \times 10^{5} \mathrm{~m} / \mathrm{s}\)
(b)
Hint: \(T=\frac{2\pi~r}{v}\)
Step: Find the time period of the electron in the first orbit.
Let be the orbital period of the electron when it is in level
Orbital period is related to orbital speed as:
where, = Radius of the orbit=\(\frac{n_{1}^{2} h^{2} \epsilon_{0}}{\pi m e^{2}}\)
h = Planck's constant = and
e = Charge on an electron
= Premitivity of free space
m = mass of an electron
\(T_{1}=\frac{2 \pi r_{1}}{v_{1}}=\frac{2 \pi \times 1^{2} \times 6.62 \times 10^{-34^{2}} \times 8.85 \times 10^{-12}}{2.18 \times 10^{6} \times \pi \times 9.1 \times 10^{-31} 1.6 \times 10^{-19^{2}}}=15.27 \times 10^{-17}=1.527 \times 10^{-16} \mathrm{sec}\)
\(T_{2}=\frac{2 \pi r_{2}}{v_{2}}={n_2}^{3} \times (T_1)=1.22\times10^{-15}~sec\)
\(T_{3}=\frac{2 \pi r_{3}}{v_{3}}={n_3}^{3} \times (T_1)=4.12\times10^{-15}~sec\)
