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NEET NEET SOLVED PAPER 2015 (C)

  • question_answer
    A certain metallic surface is illuminated with monochromatic light of wavelength \[\lambda \]. The stopping potential for photoelectric current for this light is \[3{{V}_{0}}\]. If the same surface is illuminated with light of wavelength 2\[\lambda \], the stopping potential is \[{{V}_{0}}\]. The threshold wavelength for this surface for photoelectric effect is

    A)  \[6\lambda \]                  

    B)  \[4\lambda \]                  

    C)  \[\frac{\lambda }{4}\]                  

    D)  \[\frac{\lambda }{6}\]

    Correct Answer: B

    Solution :

    From photoelectric equation \[hv+W+e{{V}_{0}}\]  (where, W = work function) So         \[\frac{hc}{\lambda }=\,W+3e{{V}_{0}}\]                                ?(i) Also,      \[\frac{hc}{2\lambda }=W+e{{V}_{0}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{hc}{\lambda }=2W+2e{{V}_{0}}\] Subtracting Eq. (i) from Eq. (ii), we get \[0=W-e{{V}_{0}}\] \[\Rightarrow \,\,\,\,W=e{{V}_{0}}\] From Eq. (i),                 \[\frac{hc}{\lambda }=e{{V}_{0}}+3e{{V}_{0}}=4e{{V}_{0}}\] The threshold wavelength is given by \[{{\lambda }_{th}}=\frac{hc}{W}=\frac{4e{{V}_{0}}\lambda }{e{{V}_{0}}}=4\lambda \]


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