JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves JEE PYQ-Photo Electric Effect X-rays

  • question_answer
    An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let \[{{\lambda }_{n}},\,\,{{\lambda }_{g}}\] be the de Broglie wavelength of the electron in the \[{{\text{n}}^{\text{th}}}\]  state and the ground state respectively. Let \[{{\Lambda }_{n}}\] be the wavelength of the emitted photon in the transition from the \[{{n}^{th}}\] state to the ground state. For large n, (A, B are constants) [JEE Main Online 08-04-2018]

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

    B) \[\Lambda _{n}^{2}\approx \lambda \]

    C) \[{{\Lambda }_{n}}\approx A+\frac{B}{\lambda _{n}^{2}}\]

    D) \[{{\Lambda }_{n}}\approx A+B{{\lambda }_{n}}\]

    Correct Answer: C

    Solution :

    [c] De Broglie wavelength\[{{\lambda }_{n}}\]
    \[=\frac{h}{mv}=\frac{h}{\frac{m{{e}^{2}}}{2n{{\in }_{0}}}}=(const)n\]
    For wavelength of emitted photon
    \[\frac{hc}{{{\Lambda }_{n}}}=13.6\left( 1-\frac{1}{{{n}^{2}}} \right)eV\]
    \[{{\Lambda }_{n}}=\frac{hc}{13.6}{{\left( 1-\frac{1}{{{n}^{2}}} \right)}^{-1}}units\]
    \[=\frac{hc}{13.6}\left( 1+\frac{1}{{{n}^{2}}} \right)units\]
    \[=A+\frac{B}{{{\lambda }_{n}}^{2}}\]
    As \[{{\lambda }_{n}}\propto n\]


You need to login to perform this action.
You will be redirected in 3 sec spinner