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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    When a piece of metal is illuminated by a monochromatic light of wavelength \[\lambda \], then stopping potential is \[3{{V}_{s}}\]. When same surface is illuminated by light of wavelength \[2\lambda \], then stopping potential becomes \[{{V}_{s}}\]. The value of threshold wavelength for photoelectric emission will be

    A)  \[4\lambda \]                                  

    B)  \[8\lambda \]

    C)  \[\frac{4}{3}\lambda \]                                                

    D)  \[6\lambda \]

    Correct Answer: A

    Solution :

    According to Einstein's photoelectric equation                                 \[eV=hc\left[ \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right]\] 1st case                                 \[3e{{V}_{s}}=hc\left[ \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right]\]       ?. (i) lInd case                                 \[e{{V}_{s}}=hc\left[ \frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}} \right]\]       ?. (ii) Dividing Eq. (i) by Eq. (ii), we get                 \[{{\lambda }_{0}}=4\lambda \]


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