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

Explanation:

$E=\frac{hc}{\lambda}$

$\lambda$ = 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:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2$

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

$\frac{1}{\lambda}=10973731.6\times \left(\frac{1}{2^2}-\frac{1}{7^2} \right )\times 1^2$

$\frac{1}{\lambda}=2.52\times 10^{6}m$

$\lambda}=3.97\times 10^{-7}m=397 nm$      $1nm=10^{-9}m$

Thus wavelength λ associated with the fifth line is 397 nm