2. Solved Papers
  3. VIT Engineering

VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    The work function of a certain metal is \[3.31\times {{10}^{-19}}J\]. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength \[\text{5000 }\overset{\text{o}}{\mathop{\text{A}}}\,\] is (Given, \[h=6.62\times {{10}^{-34}}J-s\], \[c=3\times {{10}^{8}}m{{s}^{-1}}\],\[e=1.6\times {{10}^{-19}}C\])

    A)  \[2.48\text{ }eV\]

    B)  \[0.41\text{ }eV\]

    C)  \[2.07\text{ }eV\]

    D)  \[0.82\text{ }eV\]  

    Correct Answer: B

    Solution :

    Work function \[{{W}_{0}}=3.31\times {{10}^{-19}}J\] Wavelength of incident radiation \[\lambda =5000\times {{10}^{-10}}m\] \[E={{W}_{0}}+KE\] (According to Einstein equation) \[\frac{hc}{\lambda }=3.31\times {{10}^{-19}}+KE\] \[KE=-3.31\times {{10}^{-19}}+\frac{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}\] \[=-3.31\times {{10}^{-19}}+\frac{6.62\times 3}{5}\times {{10}^{-19}}\] \[=(-3.31\times 1.324\times 3)\times {{10}^{-19}}\] \[=(3.972-3.31)\times {{10}^{-19}}=0.662\times {{10}^{-19}}J\] \[\Rightarrow \] \[E=\frac{0.662\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=0.41eV\]


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