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  • The time of revolution of an electron around a nucleus of charge  in  Bohr orbit is directly pr

The time of revolution of an electron around a nucleus of charge Ze in n^{th} Bohr orbit is directly proportional to

Option: 1

\mathrm{n}


Option: 2

\mathrm{\frac{n^{3}}{Z^{2}}}


Option: 3

\mathrm{\frac{n^{2}}{Z}}


Option: 4

\mathrm{\frac{Z}{n}}


Answers (1)

best_answer

\mathrm{T=\frac{2 \pi r}{v} ; r=\text { radius of } n^{\text {th }}orbit =\frac{n^2 h^2}{\pi m Z e^2}}
\mathrm{v=\text { speed of } e^{-} \text {in } n^{\text {th }}orbit=\frac{z e^2}{2 \varepsilon_0 n h}}
\mathrm{\therefore T=\frac{4 \varepsilon_0^2 n^3 h^3}{m Z^2 e^4} \Rightarrow T \propto \frac{n^3}{Z^2}}

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Sayak

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