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NEET AIPMT SOLVED PAPER SCREENING 2008

  • question_answer
    The work function of a surface of a photosensitive material is 6.2 eV. The wavelength, of the incident radiation for which the stopping potential is 5 V lies in the     

    A) ultraviolet region

    B) visible region

    C) infrared region 

    D) X-ray region

    Correct Answer: A

    Solution :

                   According to laws of photoelectric effect \[K{{E}_{\max }}=E-\phi \] where \[\phi \] is work function and \[K{{E}_{\max }}\]maximum kinetic energy of photoelectron. \[\therefore \]  \[hv=e{{V}_{0}}+\phi \] Or           \[hv=5eV+6.2eV=11.2eV\] \[\therefore \]  \[\lambda =\left( \frac{12400}{11.2} \right){\AA}\approx 1000{\AA}\] Hence, the radiation lies in ultraviolet region.


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