The ratio of the longest wavelengths correspo | vedclass

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

Question

(B) The wavelength $\lambda$ for a transition in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2}

Vedclass · Accepted Answer

$5/27$

Vedclass · Answer

(B) The wavelength $\lambda$ for a transition in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series,the longest wavelength corresponds to the transition from $n_2 = 2$ to $n_1 = 1$: $\frac{1}{\lambda_{Lyman}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.For the Balmer series,the longest wavelength corresponds to the transition from $n_2 = 3$ to $n_1 = 2$: $\frac{1}{\lambda_{Balmer}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36}$.Taking the ratio of the longest wavelengths: $\frac{\lambda_{Lyman}}{\lambda_{Balmer}} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.

The ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is

Similar Questions

The solution set contained in $R$ of the inequation $3^x+3^{1-x}-4 < 0$ is:

$A$ $2 \ m$ long rod of radius $1 \ cm$ which is fixed from one end is given a twist of $0.8 \ radians$. The shear strain developed will be

The component of vector $\vec{A} = 2\hat{i} + 3\hat{j}$ along the vector $\vec{B} = \hat{i} + \hat{j}$ is:

Which sugar is represented by the structural formula shown below?

Vedclass Test Series

Exam Paper Generator

Online Exam Module