Question

An electron in a hydrogen atom drops from energy level n=5 to n=3. What is the energy transition using the Rydberg equation?

Answer 1

The energy transition will be equal to `1.55 * 10^(-19)"J"`.

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

`1/(lamda) = R * (1/n_("final")^(2) - 1/n_("initial")^(2))`, where

`lamda` - the wavelength of the emitted photon;
`R` - Rydberg's constant - `1.0974 * 10^(7)"m"^(-1)`;
`n_("final")` - the final energy level - in your case equal to 3;
`n_("initial")` - the initial energy level - in your case equal to 5.

So, you've got all you need to solve for `lamda`, so

`1/(lamda) = 1.0974 * 10^(7)"m"^(-1) * (1/3^2 - 1/5^2)`

`1/(lamda) = 0.07804 * 10^(7)"m"^(-1) => lamda = 1.28 * 10^(-6)"m"`

Since `E = (hc)/(lamda)`, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by `h * c`, where

`h` - Planck's constant - `6.626 * 10^(-34)"J" * "s"`
`c` - the speed of light - `"299,792,458 m/s"`

So, the transition energy for your particular transition (which is part of the Paschen Series) is

`E = (6.626 * 10^(-34)"J" * cancel("s") * "299,792,458" cancel("m/s"))/(1.28 * 10^(-6)cancel("m"))`

`E = 1.55 * 10^(-19)"J"`

Answer 2

A photon of energy `1.531xx10^(-19)"J"` will be emitted.

The Rydberg expression for an electronic transition in the hydrogen atom is:

`1/lambda=R[(1)/n_1^(2)-(1)/n_2^(2)]`

`lambda="wavelength"`

`R=1.097xx10^(7)m^(-1)`

`n_1=3`

`n_2=5`

So:

`1/lambda=R(1/9-1/25)`

`1/lambda=R(0.071)=1.097xx10^(7)xx0.071=7.789xx10^(5)m^(-1)`

The energy of the photon given by this transition is given by:

`E=hf`

Since `c=flambda`

`E=(hc)/(lambda)`

`h` is the Planck Constant = `6.63xx10^(-34)"Js"`

`c` = the speed of light =`3xx10^(8)"m/s"`

So:

`E=6.63xx10^(-34)xx3xx10^(8)xx7.789xx10^(5)"J"`

`E=1.531xx10^(-19)"J"`