Let's solve the given problem step by step:
1. Convert the maximum energy of photoelectrons from electron volts (eV) to joules (J).
- Given:
- Maximum energy [tex]\( E_{\text{max}} = 1.2 \, \text{eV} \)[/tex]
- Charge of an electron [tex]\( e = 1.60219 \times 10^{-19} \, \text{C} \)[/tex] (Coulombs)
- Conversion:
[tex]\[
E_{\text{max}} = 1.2 \, \text{eV} \times 1.60219 \times 10^{-19} \, \text{J/eV} = 1.922628 \times 10^{-19} \, \text{J}
\][/tex]
2. Calculate the threshold frequency using the threshold wavelength.
- Given:
- Threshold wavelength [tex]\( \lambda_0 = 2.48 \times 10^{-9} \, \text{m} \)[/tex]
- Speed of light [tex]\( c = 3 \times 10^8 \, \text{m/s} \)[/tex]
- Threshold frequency [tex]\( \nu_0 \)[/tex] can be calculated using:
[tex]\[
\nu_0 = \frac{c}{\lambda_0}
\][/tex]
[tex]\[
\nu_0 = \frac{3 \times 10^8 \, \text{m/s}}{2.48 \times 10^{-9} \, \text{m}} = 1.20967742 \times 10^{17} \, \text{Hz}
\][/tex]
3. Calculate the work function (ϕ), which is the energy needed to emit the photoelectrons.
- Given:
- Planck's constant [tex]\( h = 52 \times 10^{-34} \, \text{Js} \)[/tex]
- Work function [tex]\( \phi \)[/tex] can be calculated using:
[tex]\[
\phi = h \nu_0
\][/tex]
[tex]\[
\phi = 52 \times 10^{-34} \, \text{Js} \times 1.20967742 \times 10^{17} \, \text{Hz} = 6.29032258 \times 10^{-16} \, \text{J}
\][/tex]
4. Calculate the total energy of incident light.
- The energy of the incident light [tex]\( E_{\text{incident}} \)[/tex] is the sum of the work function [tex]\( \phi \)[/tex] and the maximum energy [tex]\( E_{\text{max}} \)[/tex]:
[tex]\[
E_{\text{incident}} = \phi + E_{\text{max}} = 6.29032258 \times 10^{-16} \, \text{J} + 1.922628 \times 10^{-19} \, \text{J} = 6.29224521 \times 10^{-16} \, \text{J}
\][/tex]
5. Calculate the wavelength of the incident light.
- Using: [tex]\( E_{\text{incident}} = \frac{hc}{\lambda} \)[/tex]
- Rearrange the formula to solve for the wavelength [tex]\( \lambda \)[/tex]:
[tex]\[
\lambda = \frac{hc}{E_{\text{incident}}}
\][/tex]
[tex]\[
\lambda = \frac{52 \times 10^{-34} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{6.29224521 \times 10^{-16} \, \text{J}} = 2.47924222 \times 10^{-9} \, \text{m}
\][/tex]
Thus, the wavelength of the incident light is approximately [tex]\( 2.47 \times 10^{-9} \, \text{m} \)[/tex].
