To solve the problem, we need to calculate the minimum energy required to remove an electron from potassium metal when light of wavelength 300 nm is incident on it, given that the emitted electrons have a kinetic energy of \(1.5 \times 10^5 \, \text{J mol}^{-1}\).
### Step-by-Step Solution:
1. **Identify the given values:**
- Wavelength of light, \(\lambda = 300 \, \text{nm} = 300 \times 10^{-9} \, \text{m}\)
- Kinetic energy of emitted electrons, \(KE = 1.5 \times 10^5 \, \text{J mol}^{-1}\)
2. **Convert the kinetic energy from per mole to per electron:**
- To convert the kinetic energy from per mole to per electron, we divide by Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)):
\[
KE_{\text{per electron}} = \frac{1.5 \times 10^5 \, \text{J mol}^{-1}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 2.49 \times 10^{-19} \, \text{J}
\]
3. **Calculate the energy of the incident photon:**
- The energy of a photon can be calculated using the formula:
\[
E = \frac{hc}{\lambda}
\]
where \(h = 6.626 \times 10^{-34} \, \text{J s}\) (Planck's constant) and \(c = 3 \times 10^8 \, \text{m/s}\) (speed of light).
\[
E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{300 \times 10^{-9} \, \text{m}} \approx 6.626 \times 10^{-19} \, \text{J}
\]
4. **Use the photoelectric equation:**
- The photoelectric effect can be described by the equation:
\[
E = KE + \phi
\]
where \(E\) is the energy of the photon, \(KE\) is the kinetic energy of the emitted electron, and \(\phi\) is the work function (minimum energy needed to remove an electron from the metal).
- Rearranging gives:
\[
\phi = E - KE
\]
5. **Substitute the values to find the work function:**
\[
\phi = (6.626 \times 10^{-19} \, \text{J}) - (2.49 \times 10^{-19} \, \text{J}) \approx 4.14 \times 10^{-19} \, \text{J}
\]
### Final Answer:
The minimum energy needed to remove an electron from the potassium metal is approximately \(4.14 \times 10^{-19} \, \text{J}\).
