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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 21

  • question_answer
    If light of wavelength of maximum intensity emitted from surface at temperature T is used to cause photoelectric emission from a metallic surface, the maximum kinetic energy of the emitted electron is \[6\text{ }eV,\] which is 3 times the work function of the metallic surface. If light of wavelength of maximum intensity emitted from a surface at temperature \[{{T}_{2}}({{T}_{2}}=2{{T}_{1}})\] is used, the maximum kinetic energy of a, the photoelectrons emitted is

    A) \[2\,eV\]

    B) \[4\,eV\]   

    C) \[14\,eV\]      

    D) \[18\,eV\]

    Correct Answer: C

    Solution :

    [c] \[K{{E}_{{{\max }_{1}}}}=\frac{hC}{{{\lambda }_{1}}}-\phi \] Now \[\phi =\frac{KE{{\max }_{1}}}{3}=2eV\]  (given) So \[\frac{hC}{{{\lambda }_{1}}}=6+2=8eV\] and \[{{\lambda }_{1}}{{T}_{1}}={{\lambda }_{2}}{{T}_{2}}\] So \[\frac{hC}{{{\lambda }_{2}}}=2\frac{hC}{{{\lambda }_{1}}}=16eV\] \[K{{E}_{{{\max }_{2}}}}=\frac{hC}{{{\lambda }_{2}}}-\phi \] \[K{{E}_{\max 2}}=16-2=14eV\]            


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