What is the change in energy (in J) for an electron in a hydrogen atom when it relaxes from n=5 to n=2?

Answer 1

Hi Reese B.!

We can calculate this quantity through the use of a couple of equations. Rydberg's equation allows us to calculate the wavelength of the emitted photon by the electron as it returns to a lower energy state.

Step 1: Find the Wavelength of Emitted Photon

1/λ = RZ*(1/n_final² - 1/n_initial²)

Where:

1. R = Rydberg's constant: 1.0974x10⁷ m^-1;
2. λ = the wavelength of the photon emitted;
3. Z = the atomic mass of Hydrogen;
4. n_final = the final energy level of the electron; and
5. n_initial = the initial energy level of the electron

By plugging in our values, we get the following:

1/λ = RZ*(1/n_final² - 1/n_initial²)

=> Z = 1 because the atomic number of Hydrogen is just 1 amu

1/λ = R*(1/n_final² - 1/n_initial²)

<=> (1.0974x10⁷ m^-1)*[(1/2²) - (1/5²)]

1/λ = 2.3x10⁶ m^-1 => λ = 4.34x10^-7 m

Step 2: Find the Energy of the Emitted Photon

To calculate the energy of this emitted photon, we now must multiply Rydberg's equation by Planck's constant multiplied by the speed of light because E = hc/λ:

E = hc/λ = hc*(1/λ)

Where:

1. h = Planck's constant: 6.626x10^-34 J*s
2. c = speed of light: 299,792,458 m*s^-1

E = hc/λ

<=> (6.626x10^-34 J*s)(299,792,458 m*s^-1)/4.34x10^-7 m

E = 4.58x10^-19 J

All of our calculation has been focused on determining the energy of the photon emitted during this transition. However, by conservation of energy, we can infer that the energy emitted by the photon is equal to the energy lost by the electron. In other words, the energy emitted by the photon is the energy lost by the electron during this transition.

Hope this helps!

Cheers