Question

The kinetic energy of photoelectrons increases by 0.52 eV when the wavelength of incident light is changed from 500 nm to another web length which is approximately

• A. 1250 mm
• B. 700 nm
• C. 1000 nm
• D. 400 nm

Answer 1

The Correct Option is D

Energy of a photon is given by the equation: 

\( E = \frac{hc}{\lambda} \)

Where:

• E = energy of the photon
• h = Planck's constant = \( 6.626 \times 10^{-34} \, \text{Js} \)
• c = speed of light = \( 3 \times 10^8 \, \text{m/s} \)
• \(\lambda\) = wavelength of light

Convert the energy difference from eV to Joules:

\( 0.52 \, \text{eV} = 0.52 \times 1.6 \times 10^{-19} \, \text{J} = 8.32 \times 10^{-20} \, \text{J} \)

For the initial wavelength of 500 nm:

\( \lambda_{\text{initial}} = 500 \times 10^{-9} \, \text{m} \)

The initial energy (\( E_{\text{initial}} \)) is calculated as:

\(E_{\text{initial}} = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{500 \times 10^{-9} \, \text{m}} = 3.9768 \times 10^{-19} \, \text{J}\)

For the new wavelength \( \lambda_{\text{new}} \), the energy difference is given by:

\( E_{\text{new}} - E_{\text{initial}} = 8.32 \times 10^{-20} \, \text{J} \)

Now, solving for \( E_{\text{new}} \):

\( E_{\text{new}} = 3.9768 \times 10^{-19} + 8.32 \times 10^{-20} = 4.8 \times 10^{-19} \, \text{J} \)

Now, substitute the new energy into the equation:

\( \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{\lambda_{\text{new}}} = 4.8 \times 10^{-19} \, \text{J} \)

Solving for \( \lambda_{\text{new}} \):

\( \lambda_{\text{new}} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{4.8 \times 10^{-19}} \) \end{equation*} \[ \lambda_{\text{new}} \approx 4.140625 \times 10^{-7} \, \text{m} \]

Convert \( \lambda_{\text{new}} \) to nanometers:

\( \lambda_{\text{new}} \approx 414 \, \text{nm} \)

Therefore, the approximate new wavelength is: 414 nm

After comparing with the answer choices: The correct option is \( \boxed{400 \, \text{nm}} \), which is close to our calculated result.

Answer 2

Given the change in kinetic energy, \( \Delta K = 0.52 \, \text{eV} \) and the wavelength \( \lambda_1 = 500 \, \text{nm} \), we need to find \( \lambda_2 \). The relationship can be derived from the photoelectric effect equations:

\( K_1 = \frac{\lambda_1 hc}{\lambda_1} = \phi \) 

\( K_2 = \frac{\lambda_2 hc}{\lambda_2} - \phi \)

Now, let’s express the change in kinetic energy:

\[ K_1 - K_2 = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]

Substitute the given values for \( \Delta K = -0.52 \, \text{eV} \):

\[ - \Delta K = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \] \]

Substituting the known constants: \( h = 6.626 \times 10^{-34} \, \text{Js} \), \( c = 3.0 \times 10^8 \, \text{m/s} \), and using \( \lambda_1 = 500 \, \text{nm} \), and the value of energy:

\[ - 0.52 \, \text{eV} = (1242 \, \text{eVnm}) \left( \frac{1}{500 \, \text{nm}} - \frac{1}{\lambda_2} \right) \]

Next, solve for \( \lambda_2 \):

\[ \frac{1242}{-0.52} = 500 \left( \frac{1}{\lambda_2} \right) + \frac{1242}{0.52} \]

Simplifying the equation:

\[ \frac{1}{\lambda_2} = \frac{500}{1242} + \frac{0.52}{1242} \]

Thus, solving for \( \lambda_2 \), we get:

\[ \lambda_2 = 413 \, \text{nm} \]

Therefore, the wavelength \( \lambda_2 = 400 \, \text{nm} \).