2. Solved Papers
  3. VIT Engineering

VIT Engineering VIT Engineering Solved Paper-2008

  • question_answer
    A light whose frequency is equal to 6 x 1014 Hz is incident on a metal whose work function is \[2eV\left[ h=6.63\,\times \,{{10}^{-34}}\,Js,1eV=1.6\,\times \,{{10}^{-19}}J \right]\] The maximum energy of the electrons emitted will be

    A)  2.49 eV        

    B)  4.49 eV

    C)  0.49 eV        

    D)  5.49 eV  

    Correct Answer: C

    Solution :

    \[K{{E}_{\max }}=hv-\phi \] where hv = energy of incident photon, \[\phi \] = work function \[K{{E}_{\max }}=6.6\times {{10}^{-34}}\times 6\times {{10}^{14}}-2\times 1.6\times {{10}^{-19}}\] \[=3.96\times {{10}^{-19}}-3.2\times {{10}^{-19}}\] \[=\frac{0.76\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\] \[=0.475eV\]


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