The wavelength of the first line in the Balme | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Balmer

Vedclass · Accepted Answer

$\frac{20 \lambda}{27}$

Vedclass · Answer

(A) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$.The first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$:$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight)$.Thus,$\lambda = \frac{36}{5R}$.The second line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$:$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = R \left( \frac{4-1}{16} ight) = R \left( \frac{3}{16} ight)$.Thus,$\lambda_2 = \frac{16}{3R}$.Now,express $\lambda_2$ in terms of $\lambda$:$\lambda_2 = \frac{16}{3R} = \frac{16}{3} 	imes \left( \frac{5 \lambda}{36} ight) = \frac{16 	imes 5 \lambda}{3 	imes 36} = \frac{80 \lambda}{108} = \frac{20 \lambda}{27}$.

The wavelength of the first line in the Balmer series in the hydrogen spectrum is $\lambda$. What is the wavelength of the second line?

Similar Questions

Secondary sewage treatment is mainly a

Product $(A)$ is

The electric field for an electromagnetic wave in free space is $E = \hat{i} 30 \cos (kz - 5 \times 10^8 t)$,where the magnitude of $E$ is in $V/m$. The magnitude of the wave vector $k$ is (velocity of $EM$ wave in free space $= 3 \times 10^8 \ m/s$).

Assertion $(A):$ Above Curie temperature, a ferromagnetic material becomes paramagnetic.

Reason $(R):$ When a magnetic material is heated to very high temperature, it loses its magnetic properties.

Vedclass Test Series

Exam Paper Generator

Online Exam Module