Which line of balmer series has the shortest wavelength?
Answers
=The Balmer series or Balmer lines in atomic physics, is the designation of one of Balmer noticed that a single wavelength had a relation to every line in the hydrogen spectrum.✔✔
Answer:
For longest wavelength in Balmer series, n_2n2 = 3
Solution:
We know that,
\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]λ1=R[n121−n221]
Where,
R = constant for Rydberg's formula
\frac{1}{\lambda}=R\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]λ1=R[221−321]
\frac{1}{\lambda}=R\left[\frac{1}{4}-\frac{1}{9}\right]λ1=R[41−91]
\lambda=\frac{36}{(5 R)}λ=(5R)36
=\frac{36}{5} \times \frac{1}{R}=536×R1
=\frac{36}{5} \times 912 \dot{\mathrm{A}}=536×912A˙
=6566.4 \ \dot{\mathrm{A}}=6566.4 A˙
\lambda=656 \ \mathrm{nm}λ=656 nm
It is the longest wavelength line in balmer series.
In this question, we have made use of the Rydberg’s formula, i.e.,
\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]λ1=RZ2[n121−n221]
λ = photon wavelength,
R = constant for Rydberg's formula = 1.0973731568539 \times 10^7 m^(-1)
Z = atomic number
n_1n1 and n_2n2 are integers where n2 > n1
In the above equation, Z = 1