Question

The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is

• A. Green
• B. Blue
• C. Red
• D. Yellow

Hint:

Use the photoelectric equation to relate the energy of incident light to the work function and kinetic energy of emitted electrons. Remember that shorter wavelengths correspond to higher energy photons, which are required to overcome the work function.

Answer 1

The Correct Option is B

To ascertain the light color that can induce photoelectron emission from a metal with a work function of 3 eV, we examine the photoelectric effect. This effect describes electron emission from a metal when incident light possesses sufficient energy.

The energy \(E\) needed for electron ejection is governed by the equation:

\(E = h \cdot f\)

Where:

• \(E\) is the incident light energy (must be at least the work function),
• \(h\) is Planck's constant \((4.135667696 \times 10^{-15} \text{ eV} \cdot \text{s})\), and
• \(f\) is the light's frequency.

The work function \(W_0\) of the metal represents the minimal energy to dislodge an electron. For this problem:

\(W_0 = 3 \, \text{eV}\)

Energy is also linked to wavelength by:

\(E = \frac{h \cdot c}{\lambda}\)

Where:

• \(c\) is the speed of light \((3 \times 10^8 \, \text{m/s})\), and
• \(\lambda\) is the wavelength.

With \(E = 3 \, \text{eV}\), we calculate the wavelength \(\lambda\):

\(\lambda = \frac{h \cdot c}{E} = \frac{4.135667696 \times 10^{-15} \times 3 \times 10^8}{3}\)

Converting \(\lambda\) to nanometers:

\(\lambda = \frac{(4.135667696 \times 3 \times 10^8)}{3} \times 10^9 \approx 414 \, \text{nm}\)

This wavelength falls within the blue light range of the visible spectrum (approximately 450 nm to 495 nm). Consequently, blue light is sufficient to trigger photoelectric emission for this metal.

Therefore, the required color is: Blue.