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  • For a metal surface, the work function is 3.5 eV. If the incident light has a wavelength of 400 nm, what is the kinetic energy of the emitted electrons?  <div class='qn

For a metal surface, the work function is 3.5 eV. If the incident light has a wavelength of 400 nm, what is the kinetic energy of the emitted electrons?

 

Option: 1

2.75 eV


Option: 2

0.56 eV


Option: 3

1.75 eV

 


Option: 4

0.28 eV


Answers (1)

The photoelectric effect equation can be written as:

E_{k} = h\nu - \phi

Where E_{k} is the kinetic energy of the emitted electrons, h is Planck's constant,\nu is the frequency of the incident light, and \phiis the work function of the metal surface.

To solve this problem, we need to first calculate the frequency of the incident light using the formula:

\nu = \frac{c}{\lambda}

Where c is the speed of light and \lambda is the wavelength of the incident light.

Substituting the given values, we get:

\nu = \frac{c}{\lambda} = \frac{3 \times 10^{8} m/s}{400 \times 10^{-9} m} = 7.5 \times 10^{14} Hz

Now, substituting the values of h, \nu, and \phi in the photoelectric effect equation, we get:
E_{k} = h\nu - \phi = 6.63 \times 10^{-34} J.s \times 7.5 \times 10^{14} Hz - 3.5 eV \times 1.6 \times 10^{-19} J/eV E_{k} = 0.50 eV

Therefore, the kinetic energy of the emitted electrons is 0.50 eV, which is closest to option 2.

Posted by

Kshitij

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