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
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
To determine the stopping potential, we can use the equation for the kinetic energy of photoelectrons in terms of the stopping potential:
K.E. = eV
Where:
K.E. is the kinetic energy of the photoelectron
e is the charge of an electron (1.6 x 10^-19 C)
V is the stopping potential
Given that the maximum kinetic energy of photoelectrons is 4 eV, we can substitute into the equation:
4 eV = (1.6 x 10^-19 C) x V
To find V, we can rearrange the equation:
V = (4 eV) / (1.6 x 10^-19 C)
Now, let's calculate:
V = (4 x 1.6 x 10^-19 C) / (1.6 x 10^-19 C)
V = 4 V
Therefore, the stopping potential is 4 volts.