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Grade 11General Physics

The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

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12 Years agoGrade 11
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ApprovedApproved Tutor Answer1 Year ago

To determine whether the metal will exhibit photoelectric emission when exposed to radiation of a specific wavelength, we need to compare the energy of the incoming photons with the work function of the metal. The work function is the minimum energy required to remove an electron from the surface of the metal. In this case, the work function is given as 4.2 eV.

Calculating Photon Energy

The energy of a photon can be calculated using the formula:

E = \frac{hc}{\lambda}

Where:

  • E is the energy of the photon in joules (J).
  • h is Planck's constant, approximately 6.626 x 10^-34 J·s.
  • c is the speed of light, approximately 3.00 x 10^8 m/s.
  • λ is the wavelength of the incident radiation in meters (m).

Converting Wavelength

First, we need to convert the wavelength from nanometers to meters. Since 1 nm = 1 x 10^-9 m, we have:

λ = 330 nm = 330 x 10^-9 m

Calculating the Energy of the Photon

Now, substituting the values into the equation:

E = \frac{(6.626 x 10^{-34} J·s)(3.00 x 10^8 m/s)}{330 x 10^{-9} m}

Calculating this gives:

E ≈ 6.03 x 10^{-19} J

Converting Joules to Electronvolts

To compare this energy with the work function, we need to convert joules to electronvolts. The conversion factor is:

1 eV = 1.602 x 10^{-19} J

Thus, the energy in eV is:

E ≈ \frac{6.03 x 10^{-19} J}{1.602 x 10^{-19} J/eV} ≈ 3.77 eV

Comparing Energies

Now we have:

  • Energy of the photon: 3.77 eV
  • Work function of the metal: 4.2 eV

Since the energy of the incident photons (3.77 eV) is less than the work function (4.2 eV), the metal will not emit electrons when exposed to this radiation.

Summary

In conclusion, the metal will not exhibit photoelectric emission for incident radiation of wavelength 330 nm because the energy of the photons is insufficient to overcome the work function of the metal. This illustrates the fundamental principle of the photoelectric effect, where only photons with energy equal to or greater than the work function can cause electron emission.