NEET Sample Paper NEET Sample Test Paper-48

  • question_answer
    Light of wavelength \[0.6\mu m\]from a sodium lamp falls on a photo cell and causes the emission of photoelectrons for which the stopping potential is 0.5V with light of wavelength 0.4nm from a sodium lamp, the stopping potential is 1.5V, with this data the value of \[\frac{h}{e}\]is:

    A) \[4\times 1{{0}^{-19}}V.s\]      

    B) \[0.25\times 1{{0}^{-15}}V.s\]

    C) \[4\times 1{{0}^{-15}}V.s\]     

    D) \[4\times 1{{0}^{-8}}V.s\]

    Correct Answer: D

    Solution :

    Equation of photo electric effect \[\operatorname{W}={{W}_{o}}+e{{V}_{o}}\] \[\frac{hc}{0.6\mu m}={{W}_{o}}+e\times 0.5\]                        .....(1) \[\frac{hc}{0.4\mu m}={{W}_{o}}+0.5e\]                     .....(2) \[\left( 2 \right)-\left( 1 \right)\] \[hc\left[ \frac{1}{0.4\mu m}-\frac{1}{0.6\mu m} \right]=e\] \[hc=\frac{0.2\mu m}{0.24\mu m}=\operatorname{e}\] \[\frac{hc}{12}=\operatorname{e}\] \[\frac{h}{\operatorname{e}}=\frac{12}{c}=\frac{12}{3\times {{10}^{8}}}\] \[\frac{h}{\operatorname{e}}=4\times {{10}^{-8}}\operatorname{V}-\operatorname{s}\]


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