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  • Stuck here, help me understand: - The magnetic field associted with a light wave is given, at the origin, by . If this light falls on - Dual Nature of Matter and Radiation - JEE Main

The magnetic field associted with a light wave is given, at the origin, by B = B_{0}\left [ \sin \left ( 3.14\times 10^{7} \right ) ct+\sin \left ( 6.28\times 10^{7} \right )ct \right ]. If this light falls on a sliver plate having a work function of 4.7eV, what will be the maximum kinetic energy of the photo electrons?

\left ( c = 3\times 10^{8}ms^{-1} \right,h = 6.6\times 10^{-34}Js )

  • Option 1)

    7.72 eV

  • Option 2)

    12.5 eV

  • Option 3)

    6.82 eV

  • Option 4)

    8.52 eV

Answers (1)

best_answer

 

Conservation of energy -

h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}

h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}

 

h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}

where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function

- wherein

 

\upsilon _{1}=\frac{10^{7}}{2}C      and          \upsilon _{2}=10^{7}C

\upsilon _{2}>\upsilon _{1}

KE of photo electron will be maximum for photon with higher energy 

\upsilon _{2}=10^{7}C \: Hz

E_{ph}=h\upsilon =6.6\times 3\times 10^{-19}J

                     =12.375eV

KE_{max}=E_{ph}-\phi

                 =12.375-4.7=7.675\: eV\approx 7.7\: eV

 


Option 1)

7.72 eV

Option 2)

12.5 eV

Option 3)

6.82 eV

Option 4)

8.52 eV

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