The wavenumber of the first line (n_2=3) in t | vedclass

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

Question

(B) The Rydberg formula for wavenumber is $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the first line of the Balmer series of $H$ $(Z

Vedclass · Accepted Answer

$\frac{27 \bar{
u}_1}{5}$

Vedclass · Answer

(B) The Rydberg formula for wavenumber is $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the first line of the Balmer series of $H$ $(Z=1, n_1=2, n_2=3)$: $\bar{
u}_1 = R_H (1)^2 (\frac{1}{2^2} - \frac{1}{3^2}) = R_H (\frac{1}{4} - \frac{1}{9}) = R_H (\frac{5}{36})$.So,$R_H = \frac{36 \bar{
u}_1}{5}$.For the second line of the Balmer series of $He^{+}$ $(Z=2, n_1=2, n_2=4)$: $\bar{
u}_2 = R_H (2)^2 (\frac{1}{2^2} - \frac{1}{4^2}) = R_H (4) (\frac{1}{4} - \frac{1}{16}) = R_H (4) (\frac{3}{16}) = R_H (\frac{3}{4})$.Substituting $R_H = \frac{36 \bar{
u}_1}{5}$ into the expression for $\bar{
u}_2$: $\bar{
u}_2 = (\frac{36 \bar{
u}_1}{5}) 	imes (\frac{3}{4}) = \frac{9 	imes 3 \bar{
u}_1}{5} = \frac{27 \bar{
u}_1}{5}$.

The wavenumber of the first line $(n_2=3)$ in the Balmer series of hydrogen is $\bar{\nu}_1 \ cm^{-1}$. What is the wavenumber (in $cm^{-1}$) of the second line $(n_2=4)$ in the Balmer series of $He^{+}$?

Similar Questions

Two series of spectral lines of atomic hydrogen which do not belong to the infrared spectral region are

When an electron makes a transition from $(n+1)$ state to $n^{th}$ state,the frequency of emitted radiation is related to $n$ according to $(n >> 1)$:

The wave number of the last line of the Lyman series of the atomic spectrum of the $H$ atom corresponds to :-

$A$ quantum will have more energy if :

$A$ $12.75 \ eV$ electron beam is used to bombard gaseous hydrogen at room temperature. The wavelengths emitted belong to which of the following series?

Vedclass Test Series

Exam Paper Generator

Online Exam Module