Sure! Let's work through the problem step-by-step.
### Step 1: Understand the Problem
We are given:
- The wavelength of the incident light: [tex]\( \lambda = 2.3 \times 10^{-7} \ \text{m} \)[/tex]
- The work function of the metal: [tex]\( \phi = 3.5 \times 10^{-19} \ \text{J} \)[/tex]
- Planck's constant: [tex]\( h = 6.6 \times 10^{-34} \ \text{Js} \)[/tex]
- The speed of light: [tex]\( c = 3.0 \times 10^8 \ \text{ms}^{-1} \)[/tex]
We need to calculate the kinetic energy of the electrons that are liberated when the light is incident on the metal surface.
### Step 2: Calculate the Energy of the Incident Photons
The energy [tex]\( E \)[/tex] of a photon is given by the formula:
[tex]\[
E = \frac{hc}{\lambda}
\][/tex]
Plugging in the given values:
[tex]\[
E = \frac{(6.6 \times 10^{-34} \ \text{Js}) \cdot (3.0 \times 10^8 \ \text{ms}^{-1})}{2.3 \times 10^{-7} \ \text{m}}
\][/tex]
Calculating the numerator:
[tex]\[
(6.6 \times 10^{-34}) \times (3.0 \times 10^8) = 1.98 \times 10^{-25} \ \text{Jm}
\][/tex]
Now, divide by the wavelength:
[tex]\[
E = \frac{1.98 \times 10^{-25}}{2.3 \times 10^{-7}} \ \text{J}
\][/tex]
[tex]\[
E = 8.6086956521739125 \times 10^{-19} \ \text{J}
\][/tex]
### Step 3: Calculate the Kinetic Energy of the Liberated Electrons
The kinetic energy [tex]\( KE \)[/tex] of the electrons is given by the formula:
[tex]\[
KE = E - \phi
\][/tex]
Where [tex]\( E \)[/tex] is the energy of the incident photons and [tex]\( \phi \)[/tex] is the work function of the metal.
Plugging in the values:
[tex]\[
KE = 8.6086956521739125 \times 10^{-19} \ \text{J} - 3.5 \times 10^{-19} \ \text{J}
\][/tex]
[tex]\[
KE = 5.108695652173913 \times 10^{-19} \ \text{J}
\][/tex]
### Conclusion
The kinetic energy of the liberated electrons is:
[tex]\[
\boxed{5.108695652173913 \times 10^{-19} \ \text{J}}
\][/tex]
