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A photoelectric surface is illuminated successively by monochromatic light of wavelength \lambda and \frac{\lambda } {2} . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is :

(h = Planck's constant, c = speed of light)

  • Option 1)

    \frac{{\text{hc}}} {\lambda }

  • Option 2)

    \frac{{\text{2 hc}}} {\lambda }

  • Option 3)

    \frac{{\text{hc}}} {{3\lambda }}

  • Option 4)

    \frac{{\text{hc}}} {{2\lambda }}


Answers (1)


As discussed in

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



Let work function be \phi

K_{1}=K.E_{max}=\frac{hc}{\lambda}-\phi \:\:\:\:\:\:\:\:\:\:\:-(i)

K_{2}=\frac{hc}{\frac{\pi}{2}}-\phi=2 \frac{hc}{\lambda}-\phi \:\:\:\:\:\:\:\:\:\:\:-(ii)

According to the question, 

\frac{2hc}{\lambda}-\phi=3 \left( \frac{hc}{\lambda}- \phi \right )

Or, \frac{2hc}{\lambda}-\phi=3 \frac{3hc}{\lambda}-3 \phi

Or, \frac{2hc}{\lambda}-\phi= \frac{3hc}{\lambda}-3 \phi

Or, \phi=\frac{hc}{2\lambda}

Option 1)

\frac{{\text{hc}}} {\lambda }

This option is incorrect.

Option 2)

\frac{{\text{2 hc}}} {\lambda }

This option is incorrect.

Option 3)

\frac{{\text{hc}}} {{3\lambda }}

This option is incorrect.

Option 4)

\frac{{\text{hc}}} {{2\lambda }}

This option is correct.

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