The wavenumber of the first spectral line of | vedclass

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

Question

(A) The Rydberg formula for wavenumber is $\bar{
u} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the first spectral line of the Ly

Vedclass · Accepted Answer

$\frac{9 x}{16}$

Vedclass · Answer

(A) The Rydberg formula for wavenumber is $\bar{
u} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the first spectral line of the Lyman series of $He^{+}$ $(Z=2)$: $n_1=1, n_2=2$.$\bar{
u}_1 = R_H 	imes 2^2 	imes \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R_H 	imes 4 	imes \frac{3}{4} = 3 R_H = x$.Thus, $R_H = \frac{x}{3}$.For the second spectral line of the Balmer series of $Li^{2+}$ $(Z=3)$: $n_1=2, n_2=4$.$\bar{
u}_2 = R_H 	imes 3^2 	imes \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = R_H 	imes 9 	imes \left[ \frac{1}{4} - \frac{1}{16} ight] = R_H 	imes 9 	imes \frac{3}{16} = \frac{27 R_H}{16}$.Substituting $R_H = \frac{x}{3}$:$\bar{
u}_2 = \frac{27}{16} 	imes \frac{x}{3} = \frac{9 x}{16}$.

The wavenumber of the first spectral line of the Lyman series of $He^{+}$ ion is $x \ m^{-1}$. What is the wavenumber (in $m^{-1}$) of the second spectral line of the Balmer series of $Li^{2+}$ ion?

Similar Questions

The radii of the first,second,third,and fourth Bohr orbits of hydrogen are respectively:

Explain the Balmer series and provide its equation.

Which of the following series of transitions in the spectrum of hydrogen atom falls in the visible region?

An atom has $x$ energy levels,then the total number of spectral lines in its emission spectrum is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module