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NEET Sample Paper NEET Sample Test Paper-67

  • question_answer
    If light of wavelength a] is allowed to fall on a metal, then kinetic energy of photoelectrons emitted is\[{{E}_{1}}\]. If wavelength of light changes to \[\,{{\lambda }_{2}}\] then kinetic energy of electrons changes to\[{{E}_{2}}\]. Then work function of the metal is

    A) \[\frac{{{E}_{1}}{{E}_{2}}({{\lambda }_{1}}-{{\lambda }_{2}})}{{{\lambda }_{1}}{{\lambda }_{2}}}\]         

    B) \[\frac{{{E}_{1}}{{\lambda }_{1}}-{{E}_{2}}{{\lambda }_{2}})}{({{\lambda }_{1}}-{{\lambda }_{2}})}\]

    C) \[\frac{{{E}_{1}}{{\lambda }_{1}}-{{E}_{2}}{{\lambda }_{2}})}{({{\lambda }_{2}}-{{\lambda }_{1}})}\]                   

    D) \[\frac{{{\lambda }_{1}}{{\lambda }_{2}}-{{E}_{1}}{{E}_{2}}}{({{\lambda }_{2}}-{{\lambda }_{1}})}\]

    Correct Answer: C

    Solution :

    \[E={{W}_{0}}+{{K}_{\max }}\] \[\Rightarrow \,\,\,\,\,\frac{hc}{{{\lambda }_{1}}}={{W}_{0}}+{{E}_{1}}\,\,and\,\,\frac{hc}{{{\lambda }_{2}}}\,={{W}_{0}}+{{E}_{2}}\] \[\Rightarrow \,\,hc={{W}_{0}}{{\lambda }_{1}}+{{E}_{1}}{{\lambda }_{1}}\,and\,\,hc={{W}_{0}}{{\lambda }_{2}}+{{E}_{2}}{{\lambda }_{2}}\] \[\Rightarrow \,\,\,{{W}_{0}}{{\lambda }_{1}}+{{E}_{1}}{{\lambda }_{1}}\,=\,\,{{W}_{0}}{{\lambda }_{2}}+{{E}_{2}}{{\lambda }_{2}}\] \[\Rightarrow \,\,\,{{W}_{0}}=\frac{{{E}_{1}}{{\lambda }_{1}}-{{E}_{2}}{{\lambda }_{2}}}{({{\lambda }_{2}}-{{\lambda }_{1}})}\]


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