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JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Matter Waves

  • question_answer
    For the Bohr's first orbit of circumference \[2\pi r\], the de-Broglie wavelength of revolving electron will be [MP PMT 1987]

    A)            \[2\pi r\]                                

    B)            \[\pi r\]

    C)            \[\frac{1}{2\pi r}\]             

    D)            \[\frac{1}{4\pi r}\]

    Correct Answer: A

    Solution :

                       \[mvr=\frac{nh}{2\pi }\] According to Bohr?s theory            Þ \[2\pi r=n\,\left( \frac{h}{mv} \right)=n\lambda \]   for \[n=1\], \[\lambda =2\pi r\]


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