v_{1} is the frequency of the series limit of | vedclass

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

Question

(A) The frequency of a spectral line is given by $v = RC \left[ \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right]$.For the series limit of the Lyman s

Vedclass · Accepted Answer

$v_{1}-v_{2}=v_{3}$

Vedclass · Answer

(A) The frequency of a spectral line is given by $v = RC \left[ \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right]$.For the series limit of the Lyman series $(n_{1}=1, n_{2}=\infty)$: $v_{1} = RC \left[ 1 - \frac{1}{\infty} \right] = RC$.For the first line of the Lyman series $(n_{1}=1, n_{2}=2)$: $v_{2} = RC \left[ 1 - \frac{1}{4} \right] = \frac{3}{4} RC$.For the series limit of the Balmer series $(n_{1}=2, n_{2}=\infty)$: $v_{3} = RC \left[ \frac{1}{4} - \frac{1}{\infty} \right] = \frac{RC}{4}$.Comparing these values,we see that $v_{1} - v_{2} = RC - \frac{3}{4} RC = \frac{1}{4} RC = v_{3}$.Therefore,$v_{1} - v_{2} = v_{3}$.

$v_{1}$ is the frequency of the series limit of the Lyman series,$v_{2}$ is the frequency of the first line of the Lyman series,and $v_{3}$ is the frequency of the series limit of the Balmer series. Then:

Similar Questions

The ratio of maximum to minimum wavelength in the Balmer series of a hydrogen atom is

Let the series limit for the Balmer series be $\lambda_{1}$ and the longest wavelength for the Brackett series be $\lambda_{2}$. Then $\lambda_{1}$ and $\lambda_{2}$ are related as:

The ratio of the longest to the shortest wavelength emitted in the Paschen series of a hydrogen atom is:

The frequencies for the series limit of the Balmer and Paschen series are $v_1$ and $v_3$ respectively. If the frequency of the first line of the Balmer series is $v_2$,then the relation between $v_1, v_2,$ and $v_3$ is:

When an electron in a hydrogen atom jumps to the innermost orbit,the radiation emitted belongs to which one of the following series?

Vedclass Test Series

Exam Paper Generator

Online Exam Module