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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2007

  • 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\[=\text{ }6.6\times {{10}^{-34}}J-s\])

    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 \[{{E}_{A}}=h{{v}_{A}}\] \[=6.6\times {{10}^{-34}}\times 1.8\times {{10}^{14}}\] \[=11.88\times {{10}^{-20}}J\] \[=\frac{11.88\times {{10}^{-20}}}{1.6\times {{10}^{-19}}}eV\] \[=0.74eV\] Similarly,   \[{{E}_{B}}=0.91eV\] 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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