The shortest wavelength in the Balmer series | vedclass

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

Question

(A) Using Rydberg's formula,$\frac{1}{\lambda} = R_H Z^2 \left[ \frac{1}{n^2} - \frac{1}{m^2} \right]$,where $R_H$ is the Rydberg constant.For the sho

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Using Rydberg's formula,$\frac{1}{\lambda} = R_H Z^2 \left[ \frac{1}{n^2} - \frac{1}{m^2} \right]$,where $R_H$ is the Rydberg constant.For the shortest wavelength in the Balmer series of a hydrogen atom $(Z=1)$: $n=2, m=\infty$. Thus,$\frac{1}{\lambda_1} = R_H (1)^2 \left[ \frac{1}{2^2} - 0 \right] = \frac{R_H}{4}$,which gives $\lambda_1 = \frac{4}{R_H}$.For the shortest wavelength in the Brackett series of a hydrogen-like atom: $n=4, m=\infty$. Thus,$\frac{1}{\lambda_2} = R_H Z^2 \left[ \frac{1}{4^2} - 0 \right] = \frac{R_H Z^2}{16}$,which gives $\lambda_2 = \frac{16}{R_H Z^2}$.Given $\lambda_1 = \lambda_2$,we have $\frac{4}{R_H} = \frac{16}{R_H Z^2}$.Simplifying,$Z^2 = \frac{16}{4} = 4$,so $Z = 2$.

The shortest wavelength in the Balmer series of a hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is:

Similar Questions

The longest wavelength associated with the Paschen series is: (Given $R_H = 1.097 \times 10^7 \ m^{-1}$)

In the Balmer series,the wavelength of the $2^{\text{nd}}$ line is $\lambda_1$ and for the Paschen series,the wavelength of the $1^{\text{st}}$ line is $\lambda_2$. Then the ratio $\lambda_1 : \lambda_2$ is:

An electron jumps from the $4^{\text{th}}$ orbit to the $2^{\text{nd}}$ orbit of a hydrogen atom. Given Rydberg's constant $R_{H}=10^7 \ m^{-1}$,calculate the frequency in $Hz$ of the emitted radiation. (Take $c=3 \times 10^8 \ m/s$)

The shortest wavelength in the Lyman series is $912 \ \text{Å}$. Then, the longest wavelength in the series must be (in $\text{Å}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module