JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Photon and Photoelectric Effect

  • question_answer
    For intensity I of a light of wavelength 5000Å the photoelectron saturation current is 0.40 \[\mu \,A\] and stopping potential is 1.36 V, the work function of metal is  [RPET 1999]

    A)            2.47 eV                                   

    B)            1.36 eV

    C)            1.10 eV                                   

    D)            0.43 eV

    Correct Answer: C

    Solution :

                       By using \[E={{W}_{0}}+{{K}_{\max }}\]                    \[E=\frac{12375}{5000}=2.475\ eV\] and \[{{K}_{\max }}=e{{V}_{0}}=1.36\ eV\]            So \[2.475={{W}_{0}}+1.36\Rightarrow {{W}_{0}}=1.1\ eV\].


You need to login to perform this action.
You will be redirected in 3 sec spinner