The threshold frequency for a certain metal is 3.3 × 10^14 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut- off voltage for the photoelectric emission.

To determine the cut-off voltage for photoelectric emission when light of a certain frequency is incident on a metal, we can use the photoelectric effect equation. This phenomenon describes how light can eject electrons from a metal surface when the light's frequency exceeds a specific threshold frequency. Let's break this down step by step.

Understanding the Photoelectric Effect

The photoelectric effect is governed by Einstein's equation:

E = hf

Where:

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

When light hits the metal, if its frequency is greater than the threshold frequency, the excess energy is converted into kinetic energy of the emitted electrons. The threshold frequency (f₀) is the minimum frequency required to eject an electron from the metal surface.

Calculating the Cut-off Voltage

The cut-off voltage (V₀) can be found using the relationship between the kinetic energy of the emitted electrons and the electric potential energy. The kinetic energy (K.E.) of the emitted electrons can be expressed as:

K.E. = hf - hf₀

Where:

• hf is the energy of the incident photons.
• hf₀ is the energy corresponding to the threshold frequency.

Since the kinetic energy of the electrons can also be expressed in terms of the cut-off voltage:

K.E. = eV₀

Where e is the elementary charge (approximately 1.602 × 10^-19 C). Setting these two expressions for kinetic energy equal gives us:

eV₀ = hf - hf₀

Rearranging this leads to:

V₀ = (hf - hf₀) / e

Plugging in the Values

Now, let's calculate the cut-off voltage using the given values:

• Threshold frequency, f₀ = 3.3 × 10^14 Hz
• Incident frequency, f = 8.2 × 10^14 Hz
• Planck's constant, h = 6.626 × 10^-34 J·s
• Elementary charge, e = 1.602 × 10^-19 C

First, calculate the energies:

hf = (6.626 × 10^-34 J·s) × (8.2 × 10^14 Hz) = 5.43 × 10^-19 J

hf₀ = (6.626 × 10^-34 J·s) × (3.3 × 10^14 Hz) = 2.19 × 10^-19 J

Now, substitute these values into the equation for V₀:

V₀ = [(5.43 × 10^-19 J) - (2.19 × 10^-19 J)] / (1.602 × 10^-19 C)

V₀ = (3.24 × 10^-19 J) / (1.602 × 10^-19 C) ≈ 2.02 V

Final Result

The cut-off voltage for the photoelectric emission when light of frequency 8.2 × 10^14 Hz is incident on the metal with a threshold frequency of 3.3 × 10^14 Hz is approximately 2.02 volts. This means that the emitted electrons will have enough energy to overcome the work function of the metal and can be collected as a current if a suitable voltage is applied.