Question

light strikes a surface with $\varphi = 3.9$ ev and photoelectrons are emitted with a kinetic energy of 1.2 ev. what is the wavelength of the incident light?

\bigcirc 240 nm
\bigcirc 1040 nm
\bigcirc 460 nm
\bigcirc 730 nm
\bigcirc 320 nm

Explanation:

Step1: Recall Photoelectric Effect Formula

The photoelectric effect equation is \( E = \varphi + K.E. \), where \( E \) is the energy of the incident photon, \( \varphi \) is the work function, and \( K.E. \) is the kinetic energy of the emitted photoelectron.

Step2: Calculate Photon Energy

Given \( \varphi = 3.9 \, \text{eV} \) and \( K.E. = 1.2 \, \text{eV} \), substitute into the formula:
\( E = 3.9 \, \text{eV} + 1.2 \, \text{eV} = 5.1 \, \text{eV} \).

Step3: Convert Energy to Wavelength

Use the formula \( E = \frac{hc}{\lambda} \), where \( hc = 1240 \, \text{eV·nm} \) (a constant for photon energy - wavelength conversion). Rearrange for \( \lambda \):
\( \lambda = \frac{hc}{E} \).
Substitute \( hc = 1240 \, \text{eV·nm} \) and \( E = 5.1 \, \text{eV} \):
\( \lambda = \frac{1240 \, \text{eV·nm}}{5.1 \, \text{eV}} \approx 243 \, \text{nm} \), which is closest to 240 nm.

Answer:

240 nm