Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is:

  • Option 1)

    < 2.8 \times 10^{ - 9} \text{m}

  • Option 2)

    \geqslant \text{2}\text{.8} \times \text{l0}^{\text{ - 9}} \text{m}

  • Option 3)

    \leqslant \text{2}\text{.8} \times \text{l0}^{\text{ - 12}} \text{m}

  • Option 4)

    < 2.8 \times 10^{ - 10} \text{m}

 

Answers (1)

As learnt 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

 

 

Maximum kinetic energy of emitted electron is:

K_{max}=\frac{hc}{\lambda}-\phi=\left(\frac{1240}{500}-2.28 \right )ev

=(2.48-2.38)ev=0.10\: ev

de Broglie wavelength of the emitted electron will be:

\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m(K.E)}}

\lambda= \frac{6.67 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31}\times 1.6 \times 10^{-20}}}

=\frac{6.67 \times 10^{-34}}{\sqrt {2.912 \times10^{-50}}}=\frac{6.67}{1.70}\times 10^{-9}

\lambda=3.9 \times 10^{-9}m\geqslant 2.8 \times 10^{-9}m


Option 1)

< 2.8 \times 10^{ - 9} \text{m}

This option is incorrect.

Option 2)

\geqslant \text{2}\text{.8} \times \text{l0}^{\text{ - 9}} \text{m}

This option is correct.

Option 3)

\leqslant \text{2}\text{.8} \times \text{l0}^{\text{ - 12}} \text{m}

This option is incorrect.

Option 4)

< 2.8 \times 10^{ - 10} \text{m}

This option is incorrect.

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