JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be:
    Given E (in eV) \[=\frac{1237}{\lambda (in\,nm)}\]
    [JEE Main 10-4-2019 Morning]

    A) 1.5 eV

    B) 4.5 eV

    C) 15.1 eV                       

    D) 3.0 eV

    Correct Answer: A

    Solution :

     \[{{K}_{\max }}=\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}\]           \[\Rightarrow {{K}_{\max }}=hc\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)\]           \[\Rightarrow {{K}_{\max }}=(1237)\left( \frac{380-260}{380\times 260} \right)\]\[=1.5eV\]                       

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