The stopping potential for the photo electrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Identify the energy levels corresponding to the transitions in hydrogen atom which will result in emission of wavelength equal to that of incident radiation for the above photoelectric effect(A) n = 3 to 1 (B) n = 3 to 2 (C) n = 2 to 1 (D) n = 4 to 1

To determine which energy level transitions in a hydrogen atom correspond to the emission of a wavelength equal to that of the incident radiation in the photoelectric effect, we first need to understand the relationship between the stopping potential, the work function, and the energy of the emitted photons. The stopping potential of 10.4 V indicates the maximum kinetic energy of the emitted photoelectrons, which can be calculated using the equation:

K.E. = e * V

Here, e is the charge of an electron (approximately 1.6 x 10^-19 coulombs) and V is the stopping potential (10.4 V). Thus, the kinetic energy of the emitted electrons is:

K.E. = 1.6 x 10^-19 C * 10.4 V = 1.664 x 10^-18 J

To convert this energy into electron volts (eV), we can use the fact that 1 eV is equal to 1.6 x 10^-19 J:

K.E. = 1.664 x 10^-18 J / (1.6 x 10^-19 J/eV) = 10.4 eV

Next, we need to consider the work function (φ) of the metal, which is given as 1.7 eV. The energy of the incident photons (E_photon) can be calculated using the equation:

E_photon = K.E. + φ

Substituting the known values:

E_photon = 10.4 eV + 1.7 eV = 12.1 eV

Now, we need to find the wavelength of the incident radiation using the energy-wavelength relationship:

E = hc/λ

Where h is Planck's constant (6.626 x 10^-34 J·s) and c is the speed of light (3 x 10⁸ m/s). Rearranging this gives us:

λ = hc/E

Substituting the values:

λ = (6.626 x 10^-34 J·s * 3 x 10⁸ m/s) / (12.1 eV * 1.6 x 10^-19 J/eV)

Calculating this will give us the wavelength of the incident radiation. However, we can also directly relate the energy of the emitted photon to the transitions in the hydrogen atom.

The energy levels in a hydrogen atom are given by the formula:

E_n = -13.6 eV / n²

To find the energy difference corresponding to the transitions, we can calculate the energy for each level:

• For n = 1: E₁ = -13.6 eV
• For n = 2: E₂ = -3.4 eV
• For n = 3: E₃ = -1.51 eV
• For n = 4: E₄ = -0.85 eV

Now, we can calculate the energy differences for the transitions:

• n = 3 to 1: ΔE = E₁ - E₃ = (-13.6) - (-1.51) = 12.09 eV
• n = 3 to 2: ΔE = E₂ - E₃ = (-3.4) - (-1.51) = 1.89 eV
• n = 2 to 1: ΔE = E₁ - E₂ = (-13.6) - (-3.4) = 10.2 eV
• n = 4 to 1: ΔE = E₁ - E₄ = (-13.6) - (-0.85) = 12.75 eV

From these calculations, we see that the transition from n = 3 to n = 1 gives an energy of approximately 12.09 eV, which is very close to the 12.1 eV we calculated for the incident photon energy. Therefore, the correct answer is:

(A) n = 3 to 1