Question

Light of wavelength 5000 A° falls on a sensitive plate with photo electric work function of 1.9 eV. The kinetic energy of the emitted photoelectron will be

• A. 0.58 eV
• B. 2.48 eV
• C. 1.24 eV
• D. 1.16 eV

Hint:

The formula \(E(eV) = \frac{12400}{\lambda(\AA)}\) is a lifesaver in exams for photoelectric effect problems. It avoids the need to use \(E = hc/\lambda\) with fundamental constants and the conversion from Joules to electron-volts, saving significant time and reducing calculation errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the wavelength of incident light and the work function of a metal. We need to find the kinetic energy of the photoelectrons emitted.

Step 2: Key Formula or Approach:
We use Einstein's photoelectric equation:
\[ KE_{max} = E_{photon} - \phi \]
where \(KE_{max}\) is the maximum kinetic energy of the emitted electron, \(E_{photon}\) is the energy of the incident photon, and \(\phi\) is the work function of the material.
The energy of a photon can be calculated from its wavelength \(\lambda\). A very useful shortcut formula for this is:
\[ E_{photon} (\text{in eV}) = \frac{12400}{\lambda (\text{in Angstroms})} \]

Step 3: Detailed Explanation:
Given values are:
- Wavelength of light, \(\lambda = 5000\) Å.
- Work function, \(\phi = 1.9\) eV.
First, calculate the energy of the incident photons in eV using the shortcut formula:
\[ E_{photon} = \frac{12400}{5000} \text{ eV} \]
\[ E_{photon} = \frac{12.4}{5} = 2.48 \text{ eV} \]
Now, use the photoelectric equation to find the maximum kinetic energy:
\[ KE_{max} = E_{photon} - \phi \]
\[ KE_{max} = 2.48 \text{ eV} - 1.9 \text{ eV} \]
\[ KE_{max} = 0.58 \text{ eV} \]

Step 4: Final Answer:
The kinetic energy of the emitted photoelectron will be 0.58 eV.