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  • Solve this problem The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 e V nm)

The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 e V nm)

Option 1)

3.09 eV

 Option 2)

1.41 eV
Option 3)

1.51 eV

 Option 4)

1.68 eV

Answers (1)

best_answer

As we 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

 

 Einsteins photoelectric equation 

eV=h{\nu}-{\phi}

\Rightarrow\ \;1.68=\frac{12400}{400}eV-{\phi}=3.1-{\phi}

    {\phi}=1.42\ eV

Correct answer is 2.


Option 1)

3.09 eV

This is an incorrect option.

Option 2)

1.41 eV

This is the correct option.

Option 3)

1.51 eV

This is an incorrect option.

Option 4)

1.68 eV

This is an incorrect option.

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perimeter

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