Question

the wavelength of the first line in balmer series is 656 nm. calculate the wavelength of the second and limiting line in the balmer series​

Answer 1

Answer:

Explanation:

The balmer series

Answer 2

The wavelength of the second and limiting line in the balmer series​ is 485 m and 364 m respectively.

Explanation:

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

= Wavelength of radiation

E= energy

1)  For first line in the balmer series , the transition will be from n= 2 to n=3.

Using Rydberg's Equation:

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

$\frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\times 1^2$

$\frac{1}{\lambda}=R_H(\frac{5}{36}$

$R_H=\frac{36}{5\times 656}=0.011$

2)  the wavelength of the second line in the balmer series , the transition  will be from n= 2 to n= 4.

$\frac{1}{\lambda}=0.011\left(\frac{1}{2^2}-\frac{1}{4^2} \right )\times 1^2$

$\frac{1}{\lambda}=0.011\left(\frac{1}{2^2}-\frac{1}{4^2} \right )\times 1^2$

$\frac{1}{\lambda}=2.06\times 10^{-3}$

${\lambda}=485nm$

3) the wavelength of the limiting line in the balmer series , the transition  will be from n= 2 to n= $\infnty$

$\frac{1}{\lambda}=0.011\left(\frac{1}{2^2}-\frac{1}{\infnty^2} \right )\times 1^2$

$\frac{1}{\lambda}=$

$\frac{1}{\lambda}=2.75\times 10^{-3}$

${\lambda}=364nm$

Learn more about balmer series

https://brainly.in/question/5695407

https://brainly.in/question/4783360