Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

To determine the threshold frequency for the photoemission of electrons from a metal surface when light of a specific frequency is incident on it, we can use the photoelectric effect equation. This phenomenon describes how light can eject electrons from a material, and it’s governed by the relationship between energy, frequency, and the work function of the metal.

Understanding the Photoelectric Effect

The photoelectric effect can be described by the equation:

E = hf

Where:

• E is the energy of the incident photons.
• h is Planck's constant (approximately 6.626 × 10^-34 J·s).
• f is the frequency of the light.

When light hits the metal surface, it can provide enough energy to overcome the work function (the minimum energy required to eject an electron). The threshold frequency (f₀) is the frequency at which the energy of the photons is equal to the work function of the metal.

Calculating the Energy of the Ejected Electrons

The kinetic energy (KE) of the ejected electrons can be calculated using the formula:

KE = 0.5mv²

Where:

• m is the mass of an electron (approximately 9.11 × 10^-31 kg).
• v is the maximum speed of the ejected electrons.

Substituting the values:

KE = 0.5 × (9.11 × 10^-31 kg) × (6.0 × 10⁵ m/s)²

Calculating this gives:

KE ≈ 1.63 × 10^-19 J

Finding the Threshold Frequency

According to the photoelectric effect, the energy of the incident photons can be expressed as:

E = KE + W

Where W is the work function, which can also be expressed in terms of the threshold frequency:

W = hf₀

Thus, we can rewrite the equation as:

hf = KE + hf₀

Rearranging gives:

hf₀ = hf - KE

Now, substituting the values:

We know the frequency of the incident light is f = 7.21 × 10¹⁴ Hz.

Now, calculate the energy of the incident photons:

E = hf = (6.626 × 10^-34 J·s) × (7.21 × 10¹⁴ Hz)

E ≈ 4.78 × 10^-19 J

Now we can find the threshold frequency:

hf₀ = 4.78 × 10^-19 J - 1.63 × 10^-19 J

hf₀ ≈ 3.15 × 10^-19 J

Now, to find f₀, we rearrange:

f₀ = E₀ / h

f₀ = (3.15 × 10^-19 J) / (6.626 × 10^-34 J·s)

f₀ ≈ 4.75 × 10¹⁴ Hz

Final Result

The threshold frequency for the photoemission of electrons from the metal surface is approximately 4.75 × 10¹⁴ Hz. This frequency represents the minimum frequency of light required to eject electrons from the surface of the metal.