Certainly! Let's solve this problem step-by-step using the given data.
1. Convert the Work Function from eV to Joules:
The work function (ϕ) for aluminium is given as 4.2 eV. To convert this to Joules:
[tex]\[
1 \text{ eV} = 1.60218 \times 10^{-19} \text{ J}
\][/tex]
Therefore,
[tex]\[
ϕ = 4.2 \text{ eV} \times 1.60218 \times 10^{-19} \text{ J/eV} = 6.729156 \times 10^{-19} \text{ J}
\][/tex]
2. Calculate the Energy of the Incident Photons:
The wavelength (λ) of the incident light is given as 2000 Å. First, convert this into meters:
[tex]\[
1 \text{ Å} = 10^{-10} \text{ meters}
\][/tex]
[tex]\[
\lambda = 2000 \times 10^{-10} \text{ meters} = 2 \times 10^{-7} \text{ meters}
\][/tex]
Now, use Planck's equation to find the energy of the photons:
[tex]\[
E = \frac{hc}{λ}
\][/tex]
where:
[tex]\[
h = 6.6 \times 10^{-34} \text{ J·s}, \quad c = 3 \times 10^{8} \text{ m/s}
\][/tex]
Substitute these values in:
[tex]\[
E = \frac{(6.6 \times 10^{-34} \text{ J·s}) \times (3 \times 10^{8} \text{ m/s})}{2 \times 10^{-7} \text{ meters}} = 9.9 \times 10^{-19} \text{ J}
\][/tex]
3. Calculate the Kinetic Energy of the Emitted Electrons:
The kinetic energy (KE) of the emitted electrons can be found by subtracting the work function from the energy of the incident photons:
[tex]\[
KE = E - ϕ
\][/tex]
[tex]\[
KE = 9.9 \times 10^{-19} \text{ J} - 6.729156 \times 10^{-19} \text{ J} = 3.170844 \times 10^{-19} \text{ J}
\][/tex]
4. Find the Potential Difference Required to Stop the Electrons:
The stopping potential (V) is related to the kinetic energy of the emitted electrons by the equation:
[tex]\[
KE = eV
\][/tex]
where [tex]\( e \)[/tex] is the charge of an electron ([tex]\( 1.60218 \times 10^{-19} \text{ C} \)[/tex]).
Rearrange to solve for [tex]\( V \)[/tex]:
[tex]\[
V = \frac{KE}{e}
\][/tex]
[tex]\[
V = \frac{3.170844 \times 10^{-19} \text{ J}}{1.60218 \times 10^{-19} \text{ C}} = 1.979081002 \text{ V}
\][/tex]
Therefore, the potential difference required to stop the emission of maximum energy electrons is approximately 2 V.
