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CET Karnataka Engineering CET - Karnataka Engineering 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\[h=6.6\times {{10}^{-34}}J-s\])

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