Question

Calculate the wavelength in nanometers of the spectral line for hydrogen that corresponds to an electron falling from the n= 6 level to the n = 3 level. What region in the spectrum does this correspond to?

Answer 1

The wavelength of the spectral line for hydrogen is 656 nm.

Step-by-step explanation:

Given that,

Final state = 3

Initial state = 6

We need to calculate the wavelength of the spectral line for hydrogen

Using formula of spectral line

`(1)/(\lambda)=R((1)/(n_(f)^2)+(1)/(n_(i)^2))`

Put the value into the formula

`(1)/(\lambda)=1.0973*10^(7)((1)/(3^2)+(1)/(6^2))`

`(1)/(\lambda)=1524027.77`

`\lambda=(1)/(1524027.77)`

`\lambda=6.561*10^(-7)\ m`

`\lambda=656\ nm`

Hence, The wavelength of the spectral line for hydrogen is 656 nm.

Answer 2

Answer: Wavelength of the spectral line for hydrogen that corresponds to an electron falling from the n= 6 level to the n = 3 level is 1100 nm.

It lies in the infra red region.

Step-by-step explanation:

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

`\lambda` = Wavelength of radiation

E= energy

Using Rydberg's Equation:

`(1)/(\lambda)=R_H\left((1)/(n_i^2)-(1)/(n_f^2) \right )* Z^2`

Where,

`\lambda` = Wavelength of radiation

`R_H` = Rydberg's Constant =
`10973731.6m^(-1)`

`n_f` = Higher energy level = 6

`n_i`= Lower energy level = 3

Z= atomic number = 1 (for hydrogen)

Putting the values, in above equation, we get

`(1)/(\lambda)=R_H\left((1)/(3^2)-(1)/(6^2) \right )* 1^2`

`\lambda}=(12)/(R_H)=(12)/(10973731.6m^(-1))=1.1* 10^(-6)m=1100nm`
`1m=10^9nm`

Thus wavelength of the spectral line for hydrogen that corresponds to an electron falling from the n= 6 level to the n = 3 level is 1100 nm.

The wavelength range for gamma rays is less than 0.001 nm, The wavelength range for ultraviolet rays is 1 nm to 400 nm, The wavelength range for visible rays is 400 nm to 750 nm and The wavelength range for infra red rays is 700 nm to 1000000 nm.

Thus the wavelength of 1100 nm corresponds to infra red region.