Question:

The amount of work it takes to remove an electron from a sheet of Nickel is 5 eV.

Can visible light eject electrons from Nickel? Explain why it can or why it cannot.

As part of your answer, find the maximum wavelength of light that could be used to eject an electron from Nickel, and discuss which part of the electromagnetic spectrum it belongs to.

Answer with Full Explanation:

Step 1: Understanding Work Function and Photon Energy

The work function (ϕ) is the minimum energy required to remove an electron from the surface of a material.

For Nickel, ϕ = 5 eV.

Step 2: Maximum Energy of Visible Light

The most energetic (shortest wavelength) visible light is violet, with wavelength:
λ = 400 nm = 4.0 × 10⁻⁷ m

The energy of a photon is given by:
E = hc / λ

Where:

• h = 6.626 × 10⁻³⁴ J·s (Planck’s constant)
• c = 3.0 × 10⁸ m/s (speed of light)
• λ = 4.0 × 10⁻⁷ m

E = (6.626 × 10⁻³⁴) × (3.0 × 10⁸) / (4.0 × 10⁻⁷) ≈ 4.97 × 10⁻¹⁹ J

Convert to electronvolts:
1 eV = 1.602 × 10⁻¹⁹ J
E ≈ 4.97 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ ≈ 3.1 eV

Step 3: Comparison with Nickel’s Work Function

Since 3.1 eV < 5 eV, even the highest energy visible photons do not have enough energy to eject electrons from Nickel.

Therefore, visible light cannot cause photoelectric emission from Nickel.

Step 4: Calculating Maximum Wavelength That Can Eject Electrons

Using the photoelectric equation:
ϕ = hc / λ → λ = hc / ϕ

In electronvolt·nanometer units, hc ≈ 1240 eV·nm
λ = 1240 / 5 = 248 nm

The light must have wavelength ≤ 248 nm to eject electrons from Nickel. This falls in the ultraviolet (UV) region of the spectrum.

Final Conclusion:

• Visible light’s highest energy (3.1 eV) is less than Nickel’s work function (5 eV)
• No electrons are ejected from Nickel under visible light
• Ultraviolet light with λ ≤ 248 nm is required for photoelectric emission