Question

A sample tube consisted of atomic hydrogen in their ground state. A student illuminated the atoms with monochromatic light, that is, light of a single wavelength. If only two separate emission lines in the visible region are observed, what is the wavelength (or wavelengths) of the incident radiation?

Answer 1

The wavelength is
`\lambda =97.3 nm`

Step-by-step explanation:

Generally the series whose emission line show on the visible spectrum is the

Balmar series so this two emission line seen on the visible spectrum could either be due to the move of electron from

`n=3 \to \ n=2`

OR

`n=4 \to n=2`

This implies that the first excitement is from
`n_i=1 \to \ n_f=4`

So the energy change due to the excitement is mathematically represented as

`\Delta E = R_H [(1)/(n_i^2) -(1)/(n_f^2) ]`

substituting values

`\Delta E= R_H [(1)/(1^2) -(1)/(4^2) ]`

`\Delta E= (15)/(16) R_H`

This energy change can also be represented as

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

So
`(15)/(16) R_H = (hc)/(\lambda)`

=>
`\lambda = (16hc)/(15 R_H)`

Where
`R_H` is the Rydberg constant with a value of
`R_H = 2.18 * 10^(-18) J.`

h is the Planck's constant with values
`h = 6.626 * 10^(-34) m^2 kg / s`

c is the speed of light with value
`c = 3.0*10^(8) \ m/s`

So

`\lambda = (16(6.626 *10^(-34))(3*10^(8)))/(15 * (2.18*10^(-18)))`

`\lambda = 9.73 *10^(-8) \ m`

`\lambda =97.3 nm`