JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer The electric field of light wave is given as \[\vec{E}={{10}^{-3}}\cos \left( \frac{2\pi x}{5\times {{10}^{-7}}}-2\pi \times 6\times {{10}^{14}}t \right)\]\[\hat{x}\frac{N}{C}.\]This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is : Given, E (in eV) \[=\frac{12375}{\lambda (in\overset{\text{o}}{\mathop{\text{A}}}\,)}\]             [JEE Main 9-4-2019 Morning]

    A) 0.48 V

    B) 2.0 V

    C) 2.48 V             

    D) 0.72 V

    Correct Answer: A

    Solution :

     \[\omega =6\times {{10}^{14}}\times 2\pi \]           \[f=6\times {{10}^{14}}\]           \[C=f\lambda \] \[\lambda =\frac{C}{f}=\frac{3\times {{10}^{8}}}{6\times {{10}^{14}}}=5000\overset{\text{o}}{\mathop{\text{A}}}\,\] energy of photon \[\Rightarrow \frac{12375}{5000}\] \[=2.475\text{ }eV\] from Einstein's equation \[K{{E}_{\max }}=E-\phi \] \[e{{V}_{s}}=E-\phi \] \[e{{V}_{s}}=2.475-2\] \[e{{V}_{s}}=0.475-2\] \[e{{V}_{s}}=0.475eV\] \[{{V}_{s}}=0.475V=0.48\]volt Option [a]                   


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