Question

For the electronic transition from n = 2 to n = 4 in the hydrogen atom. a) Calculate the energy. b) Calculate the wavelength (in nm).

Answer 1

(a) The energy is,
`4.088* 10^(-19)J`

(b) The wavelength (in nm) is,
`4.862* 10^2nm`

Explanation :

The general formula for the wavelength of spectral line emitted by a hydrogen atom, when it makes a transition from
`n_2` shell to
`n_1` shell is,

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

where,

`\lambda` = wavelength

Rydberg's Constant =
`R_H=1.097* 10^7m^(-1)`

In this,
`n_1=2` and
`n_2=4`

`(1)/(\lambda)=(1.097* 10^7)* [(1)/((2)^2)-(1)/((4)^2)]`

`\lambda=4.862* 10^(-7)m=4.862* 10^2nm`

conversion used :
`(1m=10^9nm)`

Now we have to calculate the energy.

Formula used :

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

`\lambda` = Wavelength =
`4.862* 10^(-7)m`

E = energy = ?

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

h = Planck's constant =
`6.626* 10^(-34)Js`

Now put all the given values in the above formula, we get:

`E=((6.626* 10^(-34)Js)* (3* 10^8m/s))/(4.862* 10^(-7)m)`

`E=4.088* 10^(-19)J`

Therefore, the energy is
`4.088* 10^(-19)J`