Assuming that the Bohr model is valid, what is the kinetic energy of a photoelectron liberated by a photon with a the given wavelength from a state with given Z and n?
If the available energy is insufficient to break the tether, enter 0.
λ=320 nanometers from a state with Z=1 and n = 6 :
To calculate the kinetic energy of a photoelectron liberated by a photon, we need to consider the energy difference between the initial and final states of the electron.
The energy of a photon can be calculated using the equation:
E = hc/λ
where E is the energy of the photon, h is the Planck constant (6.63 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the photon.
Given that the wavelength (λ) is 320 nanometers (nm), we need to convert it to meters:
λ = 320 nm = 320 x 10^-9 meters
Plugging this value into the equation gives us:
E = (6.63 x 10^-34 J·s)(3 x 10^8 m/s)/(320 x 10^-9 m)
E ≈ 6.17 x 10^-19 J
Now, for the Bohr model, the energy of an electron in an orbit can be calculated using the formula:
En = -13.6 eV/n²
where En is the energy of the electron in the nth orbit and n is the principal quantum number.
In this case, Z=1 (since it's hydrogen), and n=6. Plugging these values into the equation gives us:
E6 = -13.6 eV/6²
E6 = -13.6 eV/36
E6 ≈ -0.38 eV
To calculate the kinetic energy, we subtract the energy of the initial state (E6) from the energy of the photon (E):
Kinetic Energy = E - E6
Converting the electronvolt (eV) to joules (J):
1 eV = 1.6 x 10^-19 J
Substituting the values and converting eV to J:
Kinetic Energy = 6.17 x 10^-19 J - (-0.38 eV)(1.6 x 10^-19 J/eV)
Kinetic Energy ≈ 6.17 x 10^-19 J + 0.61 x 10^-19 J
Kinetic Energy ≈ 6.78 x 10^-19 J
Therefore, the kinetic energy of the photoelectron liberated by a photon with a wavelength of 320 nm from a state with Z = 1 and n = 6 is approximately 6.78 x 10^-19 J.