Question

The maximum kinetic energy of photoelectrons emitted from a surface when a photons of energy 6 EV fall on it is 4 EV.the stopping potential in volts is

Answer 1

2 V.

We can use the equation:

KEmax = hν - Φ - eVs

where KEmax is the maximum kinetic energy of the photoelectrons, h is Planck's constant, ν is the frequency of the light, Φ is the work function of the surface (the minimum energy required to remove an electron), e is the elementary charge, and Vs is the stopping potential.

We are given the energy of the photon (6 eV), and we know that the maximum kinetic energy of the photoelectrons is 4 eV. We can use these values to find the work function:

Φ = hν - KEmax
Φ = (6.63 x 10^-34 J s) * (6 x 10^14 Hz) - (4 eV * 1.6 x 10^-19 J/eV)
Φ ≈ 3.1 eV

Now we can use the equation again to find the stopping potential:

Vs = (KEmax - hν + Φ) / e
Vs = (4 eV - (6.63 x 10^-34 J s) * (6 x 10^14 Hz) + 3.1 eV) / (1.6 x 10^-19 J/eV)
Vs ≈ 2 V