Question:
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.
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.
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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.
Updated On: Apr 4, 2026
- 200 nm
- 500 nm
- 250 nm
- 300 nm
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The Correct Option is B
Solution and Explanation
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
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
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