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

  • 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 [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) 220 nm                        

    B) 1700 nm

    C) 250 nm            

    D)   2020 nm

    Correct Answer: C

    Solution :

    The minimum wavelength of emitted photons is\[\frac{hc}{\Delta E}=\frac{1240}{5.6-0.7}\simeq 250nm\]


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