Physics 2000 JAMB Past Questions - Question 41: light of energy 5ev falls on a metal of work function 3ev.and electrons are liberated the stopping potential is

Correct Answer

C

Explanation

Stopping p.d =5eV-3eV=2eVThe stopping potential can be calculated using the photoelectric effect equation, which relates the energy of incident photons to the work function and the kinetic energy of emitted electrons. The stopping potential (V) can be found using the equation:The stopping potential (V_s) can be calculated using the equation:

V_s = E - W

where E is the energy of the incident light and W is the work function of the metal.

In this case, the energy of the incident light is given as 5 eV, and the work function of the metal is given as 3 eV.

Substituting these values into the equation, we have:

V_s = 5 eV - 3 eV

Simplifying the expression, we find:

V_s = 2 eV

Therefore, the stopping potential is 2 eV.

V = E - φ

Where:
- V is the stopping potential.
- E is the energy of the incident photons.
- φ is the work function of the metal.

In this case, the energy of the incident photons is given as 5 eV, and the work function of the metal is 3 eV.

So, V = 5 eV - 3 eV = 2 eV.

Therefore, the stopping potential is 2 volts (2 eV) when light with an energy of 5 eV falls on a metal with a work function of 3 eV, and electrons are liberated.