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JEE Main & Advanced Sample Paper JEE Main Sample Paper-30

  • question_answer
    When photons of energy \[4.25\,\,eV\] strike the surface of a metal A, the ejected photoelectrons have maxi mum kinetic energy, \[{{T}_{A}}\] expressed in eV and de-Broglie wavelength \[{{\lambda }_{A}}\] . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy \[4.70\,\,eV\] is \[{{T}_{B}}=({{T}_{A}}-1.50\,\,eV)\]. If the de-Broglie wavelength of these photoelectrons is \[{{\lambda }_{B}}=2{{\lambda }_{A}}\], then

    A)  The work function of A is \[2.25\,\,eV\]

    B)  The work function of B is \[2.25\,eV\]

    C) \[{{T}_{A}}=2.75\,eV\]

    D) \[{{T}_{B}}=1.25\,eV\]

    Correct Answer: A

    Solution :

    From photoelectric equation for the two metals, we have \[4.24\,eV-{{\phi }_{A}}={{T}_{A}}=\frac{P_{A}^{2}}{2m}=\frac{{{h}^{2}}}{2m\lambda _{A}^{2}}\] and \[4.70-{{\phi }_{B}}={{T}_{B}}=\frac{{{h}^{2}}}{2m\lambda _{B}^{2}}\] \[\Rightarrow \frac{{{T}_{A}}}{{{T}_{B}}}=\left( \frac{{{\lambda }_{B}}}{{{\lambda }_{A}}} \right)=4\] Also \[{{T}_{B}}={{T}_{A}}-1.5\,eV\] On solving, we get \[{{T}_{A}}=2.00\,eV\] and \[{{T}_{B}}=0.50\,eV\] \[\therefore {{\phi }_{A}}=2.25\,eV\]and \[{{\phi }_{B}}=4.20\,eV\]


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