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JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Morning)

  • question_answer
    When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy \[{{T}_{A}}\]eV and de-Broglie wavelength\[{{\lambda }_{A}}.\] The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy \[4.50\text{ }eV\]is \[{{T}_{B}}=\left( {{T}_{A}}1.5 \right)eV.\] If the de-Broglie wavelength of these photoelectrons \[{{\lambda }_{B}}=2{{\lambda }_{A}},\] then the work function of metal B is: [JEE MAIN Held On 08-01-2020 Morning]

    A) \[1.5\text{ }eV\]

    B) \[4\text{ }eV\]

    C) \[\text{3 }eV\]

    D) \[\text{2 }eV\]

    Correct Answer: B

    Solution :

    [b] de-Broglie wavelength\[\left( \lambda  \right)\], \[mv=\frac{h}{\lambda }=p=\sqrt{2m(KE)}\] \[\therefore \lambda =\frac{h}{\sqrt{2mKE}}\] \[\therefore \frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\sqrt{\frac{{{K}_{B}}}{{{K}_{A}}}}=\sqrt{\frac{{{T}_{A}}-1.5}{{{T}_{A}}}}\] (as given) Also, \[\frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{1}{2}\] On solving \[{{T}_{A}}=2ev\] \[\therefore {{K}_{B}}={{T}_{A}}-1.5=0.5ev\] \[\therefore \] Work function of metal B is \[{{\phi }_{B}}={{E}_{B}}-{{K}_{B}}\] \[=4.5-0.5=4ev\] For A, \[{{\phi }_{A}}={{E}_{A}}-{{T}_{A}}=2ev\]


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