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An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn, λg be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let \Lambdan be the wavelength of the emitted photon in the transition from the n state to the ground state. For large n, (A, B are constants)

  • Option 1)

    \Lambda _{n}^{2}\approx \lambda

  • Option 2)

    \Lambda _{n}\approx A+\frac{B}{\lambda _{n}^{2}}

  • Option 3)

    \Lambda _{n}\approx A+B\lambda _{n}}

  • Option 4)

    \Lambda _{n}^{2}\approx A+B\lambda _{n}^{2}

 

Answers (1)

best_answer

V^{_{n}}=(\frac{e^{2}}{2\epsilon _{0}h}).\frac{1}{n}

E^{_{n}}=\o -\frac{1}{2}mV_{n}^{2}

\lambda _{n}=\frac{h}{mV_{n}}=\frac{h}{m.\frac{e^{2}}{2\epsilon _{0}nh}}=\frac{2\epsilon _{0}nh^{2}}{me^{2}} (\lambda _{n> > > \lambda _{g}})

\frac{hc}{\Lambda _{n}}=E_{n}-E_{1}=\frac{1}{2}m(v_{1}^{2}-v_{n}^{2})=\frac{1}{2}m((\frac{h}{m\lambda g}^{2})-(\frac{h}{m\lambda n}))

\frac{hc}{\Lambda _{n}}=\frac{h^{2}}{2m}(\frac{1}{\lambda _{g}^{2}}-\frac{1}{\lambda _{n}^{2}})=\frac{h^{2}}{2m\lambda _{g}^{2}}(1-\frac{\lambda _{g}^{2}}{\lambda _{n}^{2}})

\frac{\Lambda ^{0}_{n}}{hc}=\frac{2m\lambda _{g}^{2}}{h^{2}}.\frac{1}{1-\frac{\lambda _{g}^{2}}{\lambda _{n}^{2}}}

\because \frac{\lambda _{g}}{\lambda _{n}}< < 1 so using Binomial expansion

\Lambda _{n}=\frac{2mc\lambda _{g}^{2}}{h}(1+\frac{\lambda _{g}^{2}}{\lambda _{n}^{2}})\Rightarrow \Lambda _{n}=A+\frac{B}{\lambda _{n}^{2}}

 

Velocity of electron in nth orbital -

v= \left ( \frac{e^{2}}{2\epsilon_{0}h} \right )\frac{z}{n}
 

- wherein

v\: \alpha \frac{z}{n}

\frac{e^{2}}{2\epsilon_{0}h}= \frac{c}{137}

 

 

 


Option 1)

\Lambda _{n}^{2}\approx \lambda

This is incorrect

Option 2)

\Lambda _{n}\approx A+\frac{B}{\lambda _{n}^{2}}

This is correct

Option 3)

\Lambda _{n}\approx A+B\lambda _{n}}

This is incorrect

Option 4)

\Lambda _{n}^{2}\approx A+B\lambda _{n}^{2}

This is incorrect

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