JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
                If the de Broglie wavelength of the electron in \[{{n}^{th}}\] Bohr orbit in a hydrogenic atom is equal to \[1.5\pi {{a}_{0}}\](\[{{a}_{0}}\]is Bohr radius), then the value of n/z is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) 1.0                               

    B) 1.50

    C) 0.75                 

    D)   0.40

    Correct Answer: C

    Solution :

    \[2\pi {{r}_{n}}=n\lambda \] \[2\pi {{a}_{0}}\times \frac{{{n}^{2}}}{Z}=n\lambda \] \[2\pi {{a}_{0}}\times \frac{n}{Z}=1.5\pi {{a}_{0}}\] \[\frac{n}{Z}=\frac{1.5\pi {{a}_{0}}}{2\pi {{a}_{0}}}=0.75\]


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