NEET Sample Paper NEET Sample Test Paper-15

  • question_answer Given that a photon of light of wavelength 10,000 angstrom has an energy equal to 1.23 eV. When light of wavelength 5000 angstrom and intensity\[{{I}_{0}}\] falls on a photoelectric cell, the saturation current is\[0.40\times {{10}^{-6}}\]ampere and the stopping potential is volt; then the work function is:

    A)  0.43 eV                      

    B)  1.10 eV

    C)  1.36 eV                      

    D)  2.47 eV

    Correct Answer: B

    Solution :

    \[E=hv=\frac{hc}{\lambda }\]or \[\frac{E}{E'}=\frac{\lambda '}{\lambda }\] or \[\frac{E}{1.23}=\frac{10,000}{5,000}\]or \[E=2.46\,eV\] Now, \[hv-W=\frac{1}{2}m{{({{v}_{\max }})}^{2}}\]or \[hv-W=e{{V}_{s}}\] \[2.46\,eV-W=e(1.36)\]volt or \[W=(2.46-1.36)=1.1\,eV\] Hence, the correction option is .


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