Question

The clectron in a hydrogen atom, originally in level n=9, undergoes a transition to a lower level by emitting a photon

of wavelength 384 nm. What is the final level of the electron?

c = 3.00 x 10 m/s,

h = 6.626 X 10^-34J-s,

Rh= 2.179 X 10^-18

Answer 1

Step-by-step explanation:

The given data is as follows.

`n_(1)` = 9,
`n_(2)` = ?

Z for hydrogen = 1

As we know that,

Energy (E) =
`(hc)/(\lambda)`

where, h = planck's constant =
`6.626 * 10^(-34)` Js

c = speed of light =
`3 * 10^(8)` m/s

`\lambda` = wavelength

According to Reydberg's equation, we will calculate the energy emitted by the photon as follows.

`\Delta E = -2.179 * 10^(-18) J * (Z)^(2)[(1)/(n^(2)_(2)) - (1)/(n^(2)_(1))]`

or,
`(hc)/(\lambda)` =
`-2.179 * 10^(-18) J * (Z)^(2)[(1)/(n^(2)_(2)) - (1)/(n^(2)_(1))]`

Putting the given values into the above equation as follows.

`(hc)/(\lambda)` =
`-2.179 * 10^(-18) J * (Z)^(2)[(1)/(n^(2)_(2)) - (1)/(n^(2)_(1))]`

`(6.626 * 10^(-34) Js * 3 * 10^(8)m/s)/(384 * 10^(-9) m)` =
`-2.179 * 10^(-18) J * (1)^(2)[(1)/(n^(2)_(2)) - (1)/((9)^(2))]`

n = 2

Thus, we can conclude that the final level of the electron is 2.