2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    The magnetic field associated with a light wave is given, at the origin, by \[B\text{ }=\text{ }{{B}_{0}}\] \[[sin(3.14\,\times \,\,{{10}^{7}})ct+sin(6.28\,\,\times \,{{10}^{6}})ct]\] If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons? \[\left( c=3\text{ }\times \text{ }{{10}^{8}}\text{ }m{{s}^{-\,1}},\text{ }h\,\,=\,\,6.6\,\,\times \,\,{{10}^{-34}}\,J-s \right)\] [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) \[6.82\text{ }eV\]          

    B)               \[8.52\text{ }eV\]

    C) \[12.5\text{ }eV\]         

    D)               \[7.55\text{ }eV\]

    Correct Answer: D

    Solution :

    \[B={{B}_{0}}[sin(3.14\times {{10}^{7}})ct+sin(6.28\,\,\times \,\,{{10}^{7}})ct]\] \[\phi \,\,=\,\,4.7eV\] \[{{K}_{\max }}\,=\,\,\frac{hc}{\lambda }\,-\,\phi \,\,=\,\,hf\,-\,\phi \] \[=\,\,\,\frac{6.6\,\times {{10}^{-34}}\,\times \,{{10}^{7}}\,c}{1.6\,\,\times \,\,{{10}^{-19}}}\,\,-\,\phi \] \[{{K}_{\max }}\,\,=\,\,\frac{6.6\,\times {{10}^{-27}}\,\times \,3\,\times \,{{10}^{8}}}{1.6\,\,\times \,\,{{10}^{-19}}}\,\,-\,4.7\,eV\] \[=\,\,\,\frac{19.6}{1.6}\,-\,4.7\,\] \[=\text{ }12.25-4.70=7.55\text{ }eV\]


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