If the wavelength of the first member of the | vedclass

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

Question

(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \r

Vedclass · Accepted Answer

$\frac{27}{32} \lambda$

Vedclass · Answer

(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $R$ is the Rydberg constant.For the Lyman series,$n_1 = 1$.The first member corresponds to the transition from $n_2 = 2$ to $n_1 = 1$:$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4} \implies R = \frac{4}{3\lambda}$.The second member corresponds to the transition from $n_2 = 3$ to $n_1 = 1$:$\frac{1}{\lambda'} = R \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = R \left( 1 - \frac{1}{9} ight) = \frac{8R}{9}$.Substituting $R = \frac{4}{3\lambda}$ into the equation for $\lambda'$:$\frac{1}{\lambda'} = \frac{8}{9} 	imes \left( \frac{4}{3\lambda} ight) = \frac{32}{27\lambda}$.Therefore,$\lambda' = \frac{27}{32} \lambda$.

If the wavelength of the first member of the Lyman series of hydrogen is $\lambda$,what is the wavelength of the second member?

Similar Questions

What is the shortest wavelength present in the Paschen series of spectral lines (in $nm$)?

The difference between the frequencies of the second and first Paschen lines of the hydrogen atom is (where $R$ is the Rydberg constant and $c$ is the speed of light in vacuum).

The first line of the Balmer series has a wavelength of $6563 \mathring{A}$. What will be the wavelength of the first member of the Lyman series in $\mathring{A}$?

The ratio of the wavelengths of the spectral lines emitted due to transitions $3 \rightarrow 2$ and $2 \rightarrow 1$ orbits in the hydrogen atom is

The ratio of the wavelengths of the first Lyman line and the second Balmer line of the hydrogen atom is

Vedclass Test Series

Exam Paper Generator

Online Exam Module