NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #24Average lifetime of a hydrogen atom excited to n = 2 state is 10-8 s. Find the number of revolutions made by the electron on the average before it jumps to the ground state.Ans : Frequency of electron (d) is given by
`` f=\frac{m{e}^{4}}{4{{\in }_{0}}^{2}{n}^{3}{h}^{3}}``
`` ``
Time period is given by
`` T=\frac{1}{f}``
`` T=\frac{4{\in }_{0}^{2}{n}^{3}{h}^{3}}{m{e}^{4}}``
Here,
h = Planck's constant
m = Mass of the electron
e = Charge on the electron
`` {\epsilon }_{0}``= Permittivity of free space
`` \therefore T=\frac{4\times \left(8.85\times {10}^{-12}\right)\times {\left(2\right)}^{3}\times {\left(6.63\times {10}^{-34}\right)}^{3}}{\left(9.10\times {10}^{-31}\right)\times {\left(1.6\times {10}^{-16}\right)}^{4}}``
`` =12247.735\times {10}^{-19}\,\mathrm{\,s\,}``
Average life time of hydrogen, t = 10`` -``8 s
Number of revolutions is given by
`` N=\frac{t}{T}``
`` \Rightarrow N=\frac{{10}^{-8}}{12247.735\times {10}^{-19}}``
N = 8.2 `` \times `` 105 revolution
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