Certainly! Let's go through the solution step-by-step to find the value of the wavelength [tex]\(\lambda\)[/tex].
Step 1: Convert the work function from electron volts to Joules
The work function ([tex]\( \phi \)[/tex]) is given as [tex]\(3.5 \, \text{eV}\)[/tex]. We need to convert this into Joules. The conversion factor is [tex]\(1 \, \text{eV} = 1.60218 \times 10^{-19} \, \text{J}\)[/tex]:
[tex]\[
\phi = 3.5 \, \text{eV} \times 1.60218 \times 10^{-19} \, \frac{\text{J}}{\text{eV}} = 5.60763 \times 10^{-19} \, \text{J}
\][/tex]
Step 2: Calculate the kinetic energy (KE) of the ejected electrons
The maximum speed ([tex]\( v \)[/tex]) of the ejected electrons is given as [tex]\(8 \times 10^5 \, \text{m/s}\)[/tex]. The mass of the electron ([tex]\( m \)[/tex]) is [tex]\( 9.1 \times 10^{-31} \, \text{kg}\)[/tex]. The kinetic energy can be calculated using the formula:
[tex]\[
\text{KE} = \frac{1}{2} m v^2
\][/tex]
[tex]\[
\text{KE} = \frac{1}{2} \times 9.1 \times 10^{-31} \, \text{kg} \times (8 \times 10^5 \, \text{m/s})^2 = 2.912 \times 10^{-19} \, \text{J}
\][/tex]
Step 3: Calculate the total energy (E) required
The total energy required to eject the electron with the given speed is the sum of the work function ([tex]\( \phi \)[/tex]) and the kinetic energy (KE):
[tex]\[
E = \phi + \text{KE}
\][/tex]
[tex]\[
E = 5.60763 \times 10^{-19} \, \text{J} + 2.912 \times 10^{-19} \, \text{J} = 8.51963 \times 10^{-19} \, \text{J}
\][/tex]
Step 4: Calculate the wavelength ([tex]\( \lambda \)[/tex]) of the light
We know that the energy ([tex]\( E \)[/tex]) of a photon is given by:
[tex]\[
E = \frac{hc}{\lambda}
\][/tex]
where [tex]\( h \)[/tex] is Planck's constant [tex]\( (6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}) \)[/tex] and [tex]\( c \)[/tex] is the speed of light in a vacuum [tex]\( (3 \times 10^8 \, \text{m/s}) \)[/tex]. Rearranging this formula to solve for the wavelength ([tex]\( \lambda \)[/tex]):
[tex]\[
\lambda = \frac{hc}{E}
\][/tex]
Substituting the values:
[tex]\[
\lambda = \frac{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{8.51963 \times 10^{-19} \, \text{J}} = 2.3332246177357468 \times 10^{-7} \, \text{m}
\][/tex]
Thus, the wavelength [tex]\(\lambda\)[/tex] of the light is approximately [tex]\(2.33 \times 10^{-7} \, \text{meters}\)[/tex].
