Question

Question #827e8

Answer 1

`4.09 * 10^(-19)J`

Explanation:

Use the formula:
`E_n = -R_H(1/n^2)`
where :
`R_H = "Rydberg constant" = 2.18 * 10^(-18)J`,
`n = "the energy level"`

For `n = 4`:
`E_4 = -2.18 * 10^(-18)J(1/4^2) = -1.3625 *10^(-19)J`

For `n = 2`
`E_2 = -2.18 * 10^(-18)J(1/2^2) = -5.45 *10^(-19)J`

Then to find the energy of the photon, you just calculate the change in energy.

`DeltaE = E_"initial" - E_"final"`

`DeltaE = (-1.3625 *10^(-19)J) - (-5.45 *10^(-19)J)`

`DeltaE = 4.0875 * 10^(-19)J`

Since you only have 3 significant figures, the answer would be:
`4.09 * 10^(-19)J`