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

Einstein's photoelectric equation is $eV_s = \frac{hc}{\lambda} - \phi_0$, where $V_s$ is the stopping potential, $\lambda$ is the wavelength, and $\phi_0$ is the work function.
We have two conditions:
Condition 1: $e(0.710) = \frac{hc}{491} - \phi_0$. (Equation 1)
Condition 2: $e(1.43) = \frac{hc}{\lambda_2} - \phi_0$. (Equation 2)
Subtract Equation 1 from Equation 2 to eliminate the work function $\phi_0$:
$e(1.43 - 0.710) = \frac{hc}{\lambda_2} - \frac{hc}{491} = hc \left(\frac{1}{\lambda_2} - \frac{1}{491}\right)$.
$e(0.72) = hc \left(\frac{1}{\lambda_2} - \frac{1}{491}\right)$.
We can use the convenient value for $hc \approx 1240$ eV·nm. The equation becomes:
$0.72 \text{ eV} = 1240 \text{ eV·nm} \left(\frac{1}{\lambda_2} - \frac{1}{491 \text{ nm}}\right)$.
$\frac{0.72}{1240} = \frac{1}{\lambda_2} - \frac{1}{491}$.
$0.0005806 = \frac{1}{\lambda_2} - 0.0020367$.
$\frac{1}{\lambda_2} = 0.0005806 + 0.0020367 = 0.0026173$.
$\lambda_2 = \frac{1}{0.0026173} \approx 382.08$ nm.
The new wavelength is approximately 382 nm.