Question

A stream of photons having energy 3 eV each impinges on a potassium surface. The work function of potassium is 2.3 eV. The emerging photo-electrons are slowed down by a copper plate placed 5 mm away. If the potential difference between the two metal plates is 1 V, the maximum distance the electrons can move away from the potassium surface before being turned back is.

• A. 3.5 mm
• B. 1.5 mm
• C. 2.5 mm
• D. 5.0 mm

Answer 1

Answer: (A)

3.5 mm

Given, the Work function of Potassium $w$ is $2.3eV$

Energy of incident photon beam is $3eV$

Thus, the maximum Kinetic Energy of the photo electrons is $KE=3-2.3=0.7eV$

At the point of returning back, the entire Kinetic Energy is converted into Electrostatic Potential Energy (Statement 1)

The Electric field between the plates is given by $E=\frac{\Delta V}{\Delta x}=\frac{1V}{5mm}=0.2Vm{m}^{-1}$

Let the distance travelled by the electron be $x$.

Increase in Electrostatic Potential energy = $E\times x$

Thus, (from statement 1)

$e\times 0.2Vm{m}^{-1}\times x=0.7eV\Rightarrow x=3.5mm$