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

To determine whether the metal with a work function of 4.2 eV will exhibit photoelectric emission when exposed to radiation of wavelength 330 nm, we need to calculate the energy of the incident photons and compare it to the work function of the metal.

Understanding Photon Energy

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

E = h * f

where:

• E is the energy of the photon in joules (J),
• h is Planck's constant, approximately 6.626 x 10^-34 J·s, and
• f is the frequency of the radiation in hertz (Hz).

Alternatively, since we have the wavelength (λ), we can use the relationship between the speed of light (c), frequency (f), and wavelength:

c = f * λ

Rearranging this gives us:

f = c / λ

Substituting this back into the energy equation gives:

E = h * (c / λ)

Calculating the Photon Energy

Now let's calculate the energy of the photons for the given wavelength of 330 nm. First, we need to convert the wavelength from nanometers to meters:

λ = 330 nm = 330 x 10^-9 m

Using the speed of light, which is approximately 3.0 x 10^8 m/s, we can find the frequency:

f = (3.0 x 10^8 m/s) / (330 x 10^-9 m) ≈ 9.09 x 10^14 Hz

Now, substituting this frequency back into the energy equation:

E = (6.626 x 10^-34 J·s) * (9.09 x 10^14 Hz) ≈ 6.02 x 10^-19 J

Converting Energy to Electron Volts

To convert the energy from joules to electron volts (eV), we use the conversion factor 1 eV = 1.602 x 10^-19 J:

E = (6.02 x 10^-19 J) / (1.602 x 10^-19 J/eV) ≈ 3.76 eV

Comparing with the Work Function

Now we compare the energy of the incoming photons (approximately 3.76 eV) with the work function of the metal (4.2 eV). Since 3.76 eV is less than 4.2 eV, the energy of the incident radiation is insufficient to overcome the work function.

Therefore, the metal will not exhibit photoelectric emission when exposed to radiation of wavelength 330 nm.