Question

If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series emission spectrum. Calculate the wavelength λ associated with the fifth line

Answer 1

Answer: Wavelength associated with the fifth line is 397 nm

Step-by-step explanation:

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

`\lambda` = Wavelength of radiation

E= energy

For fifth line in the H atom spectrum in the balmer series will be from n= 2 to n=7.

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 = 7

`n_i`= Lower energy level = 2 (Balmer series)

Putting the values, in above equation, we get

`(1)/(\lambda)=10973731.6* \left((1)/(2^2)-(1)/(7^2) \right )* 1^2`

`(1)/(\lambda)=2.52* 10^(6)m`

`\lambda}=3.97* 10^(-7)m=397 nm`
`1nm=10^(-9)m`

Thus wavelength λ associated with the fifth line is 397 nm