Question

What is the wavenumber of the radiation emitted when a hydrogen

atom makes a transition corresponding to a change in energy of
1.634 x 10-18 J?

Answer 1

Answer: Wavenumber of the radiation emitted is
`0.08* 10^(8)m^(-1)`

Step-by-step explanation:

The relationship between wavelength and energy of the wave follows the equation:

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

where,

E = energy of the radiation =
`1.634* 10^(-18)J`

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

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

`\lambda` = wavelength of radiation = ?

Putting values in above equation, we get:

`1.634* 10^(-18)J=((6.626* 10^(-34)Js)* (3* 10^8m/s))/(\lambda)\\\\\lambda=12.16* 10^(-8)m`

`\bar {\\u}=(1)/(\lambda)=(1)/(12.16* 10^(-8))=0.08* 10^(8)m^(-1)`

Thus wavenumber of the radiation emitted is
`0.08* 10^(8)m^(-1)`