To find the value of n, we can use the energy conservation principle in the photoelectric effect. The energy of the incident photon (E_photon) is equal to the work function (W) plus the kinetic energy (K.E.) of the emitted photoelectron.
E_photon = W + K.E.
The energy of a photon can be calculated using the formula:
E_photon = hc/λ
Where:
h is the Planck's constant = 6.626 x 10^-34 J·s
c is the speed of light = 3.00 x 10^8 m/s
λ is the wavelength of the emitted light in meters.
Now, we need to find the wavelength of the emitted light. When an electron in a hydrogen atom jumps from a higher energy state (n) to a lower energy state (ground state, n=1), the energy difference is given by:
ΔE = E_n - E_1 = -13.6 eV/n^2
where E_n is the energy of the electron in state n and E_1 is the energy of the electron in the ground state (n=1).
The energy of the emitted photon (E_photon) is equal to the energy difference between these states:
E_photon = ΔE = -13.6 eV/n^2
Now, we can equate this energy to the work function (W) plus the kinetic energy of the emitted photoelectron:
E_photon = W + K.E.
E_photon = 10 eV (since the stopping potential is 10 V)
Now, we can set up the equation:
-13.6 eV/n^2 = 2.75 eV + 10 eV
Simplify the equation:
-13.6 eV/n^2 = 12.75 eV
Now, solve for n:
n^2 = 13.6 eV / 12.75 eV
n^2 ≈ 1.0667
n ≈ √1.0667
n ≈ 1.0329
Since n must be a positive integer, the value of n is approximately 1. Therefore, the electron jumps from the first excited state (n=2) to the ground state (n=1).
So, the correct answer is (D) 2.
