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J & K CET Medical J & K - CET Medical Solved Paper-2003

  • question_answer
    When light of wavelength 300 nm (nanometre) falls on a photoelectric emitter, photoelectrons are just liberated. For another emitter, however, light of 600 nm wavelength is sufficient for creating photoemission. What is the ratio of the work function of the two emitters?

    A)  1 : 2                                      

    B)  2 : 1

    C)  4 : 1                                      

    D)  1 : 4

    Correct Answer: B

    Solution :

                    The minimum energy required for the emission of photoelectrons from a metal is called the work function of that metal. Given by \[W=hv\]                                  ...(i) where v is threshold frequency and h is Plancks constant. Also,       \[v=\frac{c}{\lambda }\]                                 ...(ii) where c is speed of light and k the threshold wavelength. From Eqs. (i) and (ii), we get                 \[W=\frac{hc}{\lambda }\] Given,   \[{{\lambda }_{1}}=300nm,\,{{\lambda }_{2}}=600nm\] \[\therefore \]  \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{{{W}_{1}}}{{{W}_{2}}}\] \[\Rightarrow \]               \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{600}{300}=\frac{2}{1}\]


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