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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    Radiation, with wavelength 6561\[\overset{\text{o}}{\mathop{\text{A}}}\,\] falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of \[3\text{ }\times \text{ }{{10}^{4}}\]T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to       [JEE MAIN Held on 09-01-2020 Morning]

    A) 1.6 eV  

    B) 1.1 eV

    C) 0.8 eV  

    D) 1.8 Ev

    Correct Answer: B

    Solution :

    [b] \[\phi ={{E}_{ph}}-{{\left( KE \right)}_{\max }}\] \[p=\sqrt{2mk}\] \[\Rightarrow k=\frac{{{p}^{2}}}{2m}\] \[r=\frac{p}{eB}\] \[\Rightarrow k=\frac{{{r}^{2}}{{e}^{2}}{{B}^{2}}}{2m}\] \[=\frac{12420}{6561}-\frac{{{r}^{2}}e{{B}^{2}}}{2m}\](ln eV) \[=1.89\left( eV \right)-\frac{\left( {{10}^{-4}} \right)\left( 1.6\times {{10}^{-19}} \right)9\times {{10}^{5}}}{2\times 9.07\times {{10}^{-31}}}\] \[=\left( 1.89-0.79 \right)eV=1.1eV\]


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