Sure, let's go through the solution step by step:
1. Convert the Wavelength to Meters:
- The given wavelength is [tex]\( 5400 \text{ \AA} \)[/tex].
- To convert angstroms ([tex]\( \text{\AA} \)[/tex]) to meters (m), use the conversion factor [tex]\( 1 \text{ \AA} = 10^{-10} \text{ m} \)[/tex].
- So, [tex]\( 5400 \text{ \AA} = 5400 \times 10^{-10} \text{ m} = 5.4 \times 10^{-7} \text{ m} \)[/tex].
2. Convert the Work Function to Joules:
- The given work function is [tex]\( 1.9 \text{ eV} \)[/tex].
- To convert electron volts (eV) to joules (J), use the conversion factor [tex]\( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \)[/tex].
- So, [tex]\( 1.9 \text{ eV} = 1.9 \times 1.6 \times 10^{-19} \text{ J} = 3.04 \times 10^{-19} \text{ J} \)[/tex].
3. Calculate the Energy of the Incident Photons:
- The energy [tex]\( E \)[/tex] of a photon is given by [tex]\( E = \frac{hc}{\lambda} \)[/tex].
- [tex]\( h \)[/tex] (Planck's constant) = [tex]\( 6.62 \times 10^{-34} \text{ Js} \)[/tex],
- [tex]\( c \)[/tex] (speed of light) = [tex]\( 3 \times 10^8 \text{ m/s} \)[/tex],
- [tex]\( \lambda \)[/tex] (wavelength) = [tex]\( 5.4 \times 10^{-7} \text{ m} \)[/tex],
- So, the photon energy [tex]\( E \)[/tex] is:
[tex]\[
E = \frac{6.62 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{5.4 \times 10^{-7} \text{ m}} = 3.6777777777777773 \times 10^{-19} \text{ J}
\][/tex]
4. Calculate the Kinetic Energy of the Emitted Photoelectrons:
- The kinetic energy [tex]\( K.E. \)[/tex] of the emitted photoelectrons is given by the difference between the energy of the incident photons and the work function.
- So, [tex]\( K.E. = E - \text{Work function} \)[/tex]:
[tex]\[
K.E. = 3.6777777777777773 \times 10^{-19} \text{ J} - 3.04 \times 10^{-19} \text{ J} = 6.377777777777775 \times 10^{-20} \text{ J}
\][/tex]
5. Calculate the Stopping Potential:
- The stopping potential [tex]\( V_s \)[/tex] is given by the formula [tex]\( eV_s = K.E. \)[/tex],
- where [tex]\( e \)[/tex] (elementary charge) = [tex]\( 1.6 \times 10^{-19} \text{ C} \)[/tex],
- Rearranging for [tex]\( V_s \)[/tex], we have:
[tex]\[
V_s = \frac{K.E.}{e} = \frac{6.377777777777775 \times 10^{-20} \text{ J}}{1.6 \times 10^{-19} \text{ C}} = 0.3986111111111109 \text{ V}
\][/tex]
So the energy of the photoelectrons emitted is:
- [tex]\( 6.377777777777775 \times 10^{-20} \text{ J} \)[/tex]
And their stopping potential is approximately:
- [tex]\( 0.4 \text{ V} \)[/tex]
These are the detailed steps to arrive at the solution.
