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WB JEE Medical WB JEE Medical Solved Paper-2012

  • question_answer
    When a certain metal surface is illuminated with light of frequency v, the stopping potential for photoelectric current is \[{{V}_{0}}.\] When the same surface is illuminated by light of frequency \[\frac{v}{2},\]the stopping potential is \[\frac{{{V}_{0}}}{4}.\]The threshold frequency for photoelectric, emission is

    A)  \[\frac{v}{6}\]

    B)  \[\frac{v}{3}\]

    C)  \[\frac{2v}{3}\]

    D)  \[\frac{4v}{3}\]

    Correct Answer: B

    Solution :

     Ist  case \[hv=h{{v}_{0}}+e{{V}_{0}}\] ?(i) \[\frac{hv}{2}=h{{v}_{0}}+\frac{e{{V}_{0}}}{4}\] ?(ii) Solving Eqs. (i) and (ii), we get \[{{V}_{0}}=\frac{v}{3}\]


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