Question

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is:

• A. 309 nm
• B. 329 nm
• C. 382 nm
• D. 400 nm

Hint:

When given two sets of wavelength and stopping potential, subtracting the two photoelectric equations is an efficient way to eliminate the unknown work function and solve for the required variable. Using $hc \approx 1240$ eV·nm simplifies calculations when energy is in eV and wavelength is in nm.

Answer 1

The Correct Option is C

The problem is based on the photoelectric effect, which relates to the energy of electrons emitted from a surface when light of a certain wavelength falls on it. The stopping potential (\(V_0\)) is the potential required to stop the most energetic photoelectrons emitted due to the light incident on the surface. 

According to Einstein's photoelectric equation, the energy of the incident light is related to the stopping potential and the work function of the material:

\(E = \text{hf} = \text{h}\frac{c}{\lambda}\), where:

• \(E\) is the energy of the incident photons,
• \(\text{h}\) is Planck's constant (\(6.626 \times 10^{-34} \ \text{Js}\)),
• 3 \times 10^8 \ \text{m/s}),
• \(\lambda\) is the wavelength of the incident light.

Energy for the photoelectric effect can also be expressed as:

\(\text{E} = \text{eV}_0 + \phi\)

Where:

• \(\text{e}\) is the charge of an electron,
• \phi is the work function.

Now, let's solve the problem using this step-by-step approach:

1. First, calculate the initial energy with \(\lambda = 491 \ \text{nm}\) and \text{E}_1 = \text{h}\frac{c}{\lambda_1} = \text{eV}_{01} + \phi

 

1. For the new wavelength and stopping potential, the equation modifies to:

\(\text{E}_2 = \text{h}\frac{c}{\lambda_2} = \text{eV}_{02} + \phi\)

Where:

• \phi cancels when calculating the difference between two states.

1. Now, derive the expression to find the new wavelength \text{h}\frac{c}{\lambda_1} - \text{h}\frac{c}{\lambda_2} = \text{e}(\text{V}_{02} - \text{V}_{01})

 

This can be rearranged to solve for \frac{1}{\lambda_2} = \frac{1}{\lambda_1} + \frac{\text{e(V}_{02} - \text{V}_{01})}{\text{hc}}

1. Calculation:

Substitute the values:

\lambda_2 \approx 382 \ \text{nm}.

Thus, the new wavelength is 382 nm.