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 longest wavelength of light that can cause the emission of photoelectrons from a substance, we apply the photoelectric effect principle. The photoelectric effect equation is as follows:

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

Where:

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

The work function (\(\phi\)) of the substance is given as 3.0 eV, which is the minimum energy needed to emit photoelectrons. For the longest wavelength:

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

Convert the energy from electron volts to joules for computation:

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

Now, substitute the values into the equation:

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

Solve for \(\lambda_{\text{max}}\):

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

Calculate this value:

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

The longest wavelength that can cause the emission of photoelectrons from this substance is therefore approximately 414 nm.

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

Answer 2

For photoelectric emission, the energy of the photon must be equal to or greater than the work function \( W_e \):

\[ \lambda = \frac{hc}{W_e} \]

Using \( h = 1240 \, \text{nm} \times \text{eV} \) and \( W_e = 3.0 \, \text{eV} \):

\[ \lambda \leq \frac{1240 \, \text{nm} \times \text{eV}}{3.0 \, \text{eV}} = 413.33 \, \text{nm} \]

Thus, \( \lambda_{\text{max}} \approx 414 \, \text{nm} \).