Question

The work function of a substance is 3.0 eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:

• A. 215 nm
• B. 400 nm
• C. 414 nm
• D. 200 nm

Answer 1

The Correct Option is C

To determine the maximum wavelength of light capable of eliciting photoelectrons from a substance, we employ the photoelectric effect principle, governed by the equation:

\(E = h \cdot u = \dfrac{h \cdot c}{\lambda}\)

Definitions of terms used:

• \(E\) represents the energy of the incident photon (measured in electron volts, eV).
• \(h\) is Planck's constant, with a value of approximately \(6.626 \times 10^{-34} \, \text{Js}\).
• \(u\) denotes the frequency of the incident light.
• \(c\) is the speed of light in a vacuum, which is \(3.0 \times 10^8 \, \text{m/s}\).
• \(\lambda\) indicates the wavelength of the incident light (in meters).

The substance's work function (\(\phi\)) is provided as 3.0 eV, representing the minimum energy required for photoelectron emission. The relationship for the longest wavelength is:

\(\phi = \dfrac{h \cdot c}{\lambda_{\text{max}}}\)

Convert the energy from electron volts to joules for calculation:

\(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\)
\(\phi = 3.0 \, \text{eV} = 3.0 \times 1.602 \times 10^{-19} \, \text{J} = 4.806 \times 10^{-19} \, \text{J}\)

Insert the known values into the equation:

\(4.806 \times 10^{-19} = \dfrac{6.626 \times 10^{-34} \cdot 3.0 \times 10^8}{\lambda_{\text{max}}}\)

Rearrange to solve for \(\lambda_{\text{max}}\):

\(\lambda_{\text{max}} = \dfrac{6.626 \times 10^{-34} \cdot 3.0 \times 10^8}{4.806 \times 10^{-19}}\)

The computed value is:

\(\lambda_{\text{max}} \approx 4.14 \times 10^{-7} \, \text{m} = 414 \, \text{nm}\)

Consequently, the longest wavelength capable of inducing photoelectron emission from this substance is approximately 414 nm.

Therefore, the correct answer is 414 nm, corresponding to option \(\text{414 nm}\).