Final answer:
The energy levels corresponding to the transitions in a hydrogen atom which will result in emission of wavelength equal to that of the incident radiation for the given photoelectric effect are n = 2 to 1.
Step-by-step explanation:
The stopping potential depends on the difference between the energy of the incident photon and the work function of the metal. In this case, the stopping potential is 10.4 V, which corresponds to the maximum kinetic energy of the photoelectrons. The work function of the metal is 1.7 eV.
The energy of a photon is given by the equation E = hf, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), and f is the frequency of the electromagnetic wave.
The energy of a photon can also be calculated using the equation E = hc/λ, where c is the speed of light (3.00 x 10^8 m/s) and λ is the wavelength of the electromagnetic wave.
To emit photoelectrons with a wavelength equal to that of the incident radiation, the energy of the photon should be equal to the work function of the metal.
We can calculate the energy of a photon using the equation E = hf and find the corresponding wavelength using the equation E = hc/λ.
We need to find the energy levels of the hydrogen atom transitions that result in a photon with energy equal to 1.7 eV.
By calculating the energy of the transitions from one energy level to another using the Rydberg formula: E = -13.6 eV * (1/n1^2 - 1/n2^2), where n1 and n2 are the initial and final energy levels, we can determine the transitions that result in a photon with energy equal to 1.7 eV or a wavelength equal to that of the incident radiation.
Using the Rydberg formula, we can calculate the energy transitions for each option:
n = 3 to 1:
E = -13.6 eV * (1/3^2 - 1/1^2)
= -13.6 eV * (1/9 - 1)
= 13.6 eV * (8/9)
= 12.09 eV
n = 4 to 1:
E = -13.6 eV * (1/4^2 - 1/1^2)
= -13.6 eV * (1/16 - 1)
= 13.6 eV * (15/16)
= 12.90 eV
n = 2 to 1:
E = -13.6 eV * (1/2^2 - 1/1^2)
= -13.6 eV * (1/4 - 1)
= 13.6 eV * (3/4)
= 10.20 eV
Based on these calculations, only option C (n = 2 to 1) corresponds to a transition that results in a photon with energy equal to the work function of the metal (1.7 eV).
Therefore, the energy levels corresponding to the transitions in a hydrogen atom which will result in the emission of a wavelength equal to that of the incident radiation are n = 2 to 1.