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KVPY Sample Paper KVPY Stream-SX Model Paper-6

  • question_answer
    In a Frank-hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

    A) 1700 nm                       

    B) 2020 nm

    C)  220 nm           

    D) 250 nm

    Correct Answer: D

    Solution :

    \[5.6eV-0.7eV=4.9eV\]
    \[=\frac{12410eV-A{}^\circ }{\lambda }\]
    \[=\frac{12410eV-A{}^\circ }{\lambda }\]
    \[\lambda =\frac{12410eV-A{}^\circ }{4.9eV}\approx 250nm.\]


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