Question

According to the Bohr model for the hydrogen atom, how much energy is necessary to excite an electron from n=1 to n=2?

Answer 1

This energy can be determined from the Rydberg equation, which is

`\mathbf(1/lambda = R_H(1/(n_i^2) - 1/(n_j^2)))`

where:

• `lambda` is the wavelength in units of `"m"`.
• `R_H` is the Rydberg constant, `"10973731.6 m"^-1`.
• `n_i` is the principal quantum number `n` for the lower energy level.
• `n_j` is the principal quantum number `n` for the higher energy level.

In this case, `n_i = 1` and `n_j = 2`, since the excitation from `n = 1` to `n = 2` is upwards. Thus:

`1/lambda = "10973731.6 m"^-1(1/1^2 - 1/2^2)`

`= "10973731.6 m"^-1(1/1^2 - 1/2^2)`

`= "8230298.7 m"^(-1)`

Now, the wavelength is

`color(green)(lambda) = 1/("8230298.7 m"^(-1))`

`= color(green)(1.215xx10^(-7) "m"),`

or `"121.5 nm"`. Let's relate this back to the energy. Recall the equation:

`\mathbf(DeltaE = hnu = (hc)/lambda)`

where:

• `DeltaE` is the change in energy in `"J"`.
• `h` is Planck's constant, `6.626xx10^(-34) "J"cdot"s"`.
• `nu` is the frequency in `"s"^(-1)`. Remember that `c = lambdanu`.
• `c` is the speed of light, `2.998xx10^8 "m/s"`.
• `lambda` is the wavelength in `"m"` like before.

So, the absorption of energy into a single hydrogen atomic system associated with this process is:

`color(blue)(DeltaE) = ((6.626xx10^(-34) "J"cdotcancel"s")(2.998xx10^(8) cancel"m/s"))/(1.215xx10^(-7) cancel"m")`

`= color(blue)(1.635xx10^(-18) "J")`

(absorption is an increase in energy for the system, thus `DeltaE > 0`.)