To determine the threshold wavelength of the metal for which the work function is given, we first need to apply the relevant physics concepts involving the photoelectric effect.
- The work function (\(\phi\)) is the minimum energy required to remove an electron from the surface of a metal. This is related to the threshold frequency (\(\nu_0\)) by the equation:
\(\phi = h\cdot \nu_0\) - The relationship between frequency (\(\nu\)) and wavelength (\(\lambda\)) is given by the equation:
\(\nu = \frac{c}{\lambda}\), where \(c\) is the speed of light. - Substituting frequency into the work function expression, we get:
\(\phi = \frac{h\cdot c}{\lambda_0}\) - Rearranging the formula to find the threshold wavelength (\(\lambda_0\)), we get:
\(\lambda_0 = \frac{h \cdot c}{\phi}\) - Substituting given values:
- Work function, \(\phi = 6.875\) eV. To convert this to joules (since \(h\) uses joules), multiply by charge of an electron: \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).
- \(\phi = 6.875 \times 1.6 \times 10^{-19} \text{ J} = 1.1 \times 10^{-18} \text{ J}\)
- Planck's constant, \(h = 6.626 \times 10^{-34} \text{ Js}\)
- Speed of light, \(c = 3 \times 10^8 \text{ m/s}\)
- Substituting these into the wavelength formula:
\(\lambda_0 = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^{8}}{1.1 \times 10^{-18}}\)
\(\lambda_0 = \frac{1.9878 \times 10^{-25}}{1.1 \times 10^{-18}}\)
\(\lambda_0 \approx 1.806 \times 10^{-7} \text{ m}\) - Convert the wavelength into Angstroms:
\(\lambda_0 = 1.806 \times 10^{-7} \text{ m} \times 10^{10} \text{ Å/m}\)
\(\lambda_0 \approx 1806 \text{ Å}\)
Therefore, the threshold wavelength of the metal is approximately 1800 Å, which matches the correct answer provided in the options.