Work function for an object is 2.3 eV. If maximum kinetic energy of ejected electrons is 0.18 eV, find wavelength \( \lambda \) of incident photon on object.
Show Hint
Remember, the photoelectric equation relates the energy of the photon to the work function and the kinetic energy of ejected electrons. Use \( E_{\text{photon}} = \frac{h c}{\lambda} \) to find the wavelength.
Step 1: Understanding the photoelectric equation.
The energy of a photon is given by the equation:
\[
E_{\text{photon}} = \frac{h c}{\lambda}
\]
Where:
- \( h \) is Planck's constant \( = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \),
- \( c \) is the speed of light \( = 3 \times 10^8 \, \text{m/s} \),
- \( \lambda \) is the wavelength of the incident photon.
The photoelectric equation is given by:
\[
E_{\text{photon}} = \text{Work Function} + E_{\text{kinetic}}
\]
Where:
- The Work Function \( \phi = 2.3 \, \text{eV} \),
- The maximum kinetic energy \( E_{\text{kinetic}} = 0.18 \, \text{eV} \).
Step 2: Finding the total energy of the photon.
Substituting the given values into the photoelectric equation:
\[
E_{\text{photon}} = 2.3 \, \text{eV} + 0.18 \, \text{eV} = 2.48 \, \text{eV}
\]
Step 3: Converting the energy to joules.
Since \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \), we convert the photon energy to joules:
\[
E_{\text{photon}} = 2.48 \, \text{eV} = 2.48 \times 1.602 \times 10^{-19} \, \text{J} = 3.97 \times 10^{-19} \, \text{J}
\]
Step 4: Finding the wavelength.
Now, using the energy-wavelength relation \( E_{\text{photon}} = \frac{h c}{\lambda} \), we solve for \( \lambda \):
\[
\lambda = \frac{h c}{E_{\text{photon}}}
\]
Substituting the known values:
\[
\lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{3.97 \times 10^{-19}} = 5.00 \times 10^{-7} \, \text{m} = 500 \, \text{nm}
\]
Step 5: Conclusion.
Therefore, the wavelength of the incident photon is \( \lambda = 500 \, \text{nm} \).
Final Answer: 500 nm
