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MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • question_answer
    A and B are two metals with threshold frequencies \[1.8\times {{10}^{14}}Hz\]and \[2.2\times {{10}^{14}}Hz.\] Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by \[(Take\,h=6.6\times {{10}^{-34}}J-s)\]

    A) B alone                                

    B) A alone

    C) Neither A nor B                

    D) Both A and B

    Correct Answer: B

    Solution :

    Threshold energy of A is \[{{E}_{A}}=h{{v}_{A}}\]                 \[=6.6\times {{10}^{-34}}\times 1.8\times {{10}^{14}}\]                 \[=11.88\times {{10}^{-20}}J\]                 \[=\frac{11.88\times {{10}^{-20}}}{1.6\times {{10}^{-19}}}eV\]                 \[=0.74\text{ }eV\] Similarly,  \[{{E}_{B}}=0.91\text{ }eV\] Since, the incident photons have energy greater than\[{{E}_{A}}\]but less than\[{{E}_{B}}\]. So, photoelectrons will be emitted from metal A only.


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