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  • Tell me? When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. the work functionof the metal is:

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. the work functionof the metal is:

  • Option 1)

    0.65 eV

  • Option 2)

    1.0 eV

  • Option 3)

    1.3 eV

  • Option 4)

    1.5 eV

 

Answers (1)

best_answer

As we discussed in concept

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 is \phi

Initial energy of incidence = E

=> Kinetic energy = E-\phi = 0.5eV-----(1)

Finally, 1.2\:E- \phi=0.8eV------(2)

From (1) E=0.5eV+ \phi

=> 1.2(0.5eV+ \phi)\:-\: \phi=0.8eV

=> 0.6+0.2 \phi = 0.8eV\:\:\:=> \phi = 1eV


Option 1)

0.65 eV

This option is incorrect.

Option 2)

1.0 eV

This option is correct.

Option 3)

1.3 eV

This option is incorrect.

Option 4)

1.5 eV

This option is incorrect.

Posted by

prateek

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