Question

The work function of a metal is \(4.0\text{ eV}\). If the metal is irradiated with radiation of wavelength \(200\text{ nm}\), the maximum kinetic energy of the photoelectrons would be about:
(Use \(hc = 1240\text{ eV}\cdot\text{nm}\))

• A. \(6.2\text{ eV}\)
• B. \(4.0\text{ eV}\)
• C. \(2.2\text{ eV}\)
• D. \(8.2\text{ eV}\)

Hint:

Using \(E = \frac{1240}{\lambda \text{ (in nm)}} \text{ eV}\) or \(E = \frac{12400}{\lambda \text{ (in \AA)}} \text{ eV}\) is a massive time-saver for calculating photon energy in Modern Physics problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to calculate the maximum kinetic energy of the emitted photoelectrons when a metal surface with a known work function is exposed to light of a specific wavelength.

Step 2: Key Formula or Approach:
According to Einstein's photoelectric equation, the maximum kinetic energy (\(K_{\max}\)) of the emitted electrons is given by:
\[ K_{\max} = E - \Phi \] where \(E\) is the energy of the incident photon and \(\Phi\) is the work function of the metal.
The energy of a photon can be calculated using the relation:
\[ E = \frac{hc}{\lambda} \]

Step 3: Detailed Explanation:
Given values are:
Wavelength of incident radiation, \(\lambda = 200\text{ nm}\).
Work function of the metal, \(\Phi = 4.0\text{ eV}\).
Using the provided constant \(hc = 1240\text{ eV}\cdot\text{nm}\), we first find the energy of the incident photon:
\[ E = \frac{1240\text{ eV}\cdot\text{nm}}{200\text{ nm}} = 6.2\text{ eV} \] Now, substitute the values into the photoelectric equation to find the maximum kinetic energy:
\[ K_{\max} = 6.2\text{ eV} - 4.0\text{ eV} = 2.2\text{ eV} \]

Step 4: Final Answer:
The correct choice is (C).