calculate the energies of the states of the hydrogen atom with n=2 and n=3, and calculate the wavelength of the photon emitted by the atom when an electron makes a transition between these two states.
See your other post above.
https://www.jiskha.com/questions/1779568/what-is-the-wavelength-of-a-photon-emitted-during-a-transition-from-the-initial-state-ni-5
To calculate the energies of the states of the hydrogen atom with n=2 and n=3, we can use the equation for the energy of an electron in the hydrogen atom:
E = -13.6 eV / n^2
where E is the energy in electron volts (eV) and n is the principal quantum number.
For n=2:
E2 = -13.6 eV / 2^2 = -13.6 eV / 4 = -3.4 eV
For n=3:
E3 = -13.6 eV / 3^2 = -13.6 eV / 9 = -1.51 eV
Next, to calculate the wavelength of the photon emitted when an electron transitions from n=3 to n=2, we can use the following equation:
ΔE = E2 - E3
where ΔE is the change in energy between the two states. Since the electron is transitioning from a higher energy state (n=3) to a lower energy state (n=2), the change in energy is negative.
ΔE = -3.4 eV - (-1.51 eV) = -1.89 eV
To find the wavelength, we can use the equation:
λ = hc / ΔE
where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and ΔE is the change in energy in joules (J).
First, we need to convert the energy difference from eV to J. The conversion factor is 1 eV = 1.602 x 10^-19 J.
ΔE = -1.89 eV * (1.602 x 10^-19 J/eV) = -3.03 x 10^-19 J
Now we can calculate the wavelength:
λ = (6.626 x 10^-34 J·s) * (3.00 x 10^8 m/s) / (-3.03 x 10^-19 J)
λ = 6.55 x 10^-7 m
Therefore, the wavelength of the photon emitted by the hydrogen atom when an electron transitions from n=3 to n=2 is approximately 6.55 x 10^-7 meters (655 nm).