The wave number of the first line of the Balm | vedclass

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

Question

(C) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^

Vedclass · Accepted Answer

$136800$

Vedclass · Answer

(C) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$ and for the first line,$n_2 = 3$.For hydrogen $(H)$,$Z = 1$,so $\bar{
u}_H = R(1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 15200 \, cm^{-1}$.For $Li^{2+}$ ion,$Z = 3$. The wave number for the first line of the Balmer series is $\bar{
u}_{Li^{2+}} = R(3)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight)$.Substituting the value of $R \left( \frac{1}{2^2} - \frac{1}{3^2} ight)$ from the hydrogen equation:$\bar{
u}_{Li^{2+}} = 3^2 	imes \bar{
u}_H = 9 	imes 15200 = 136800 \, cm^{-1}$.

The wave number of the first line of the Balmer series in the hydrogen spectrum is $15200 \, cm^{-1}$. What will be the wave number of the first line of the Balmer series for the $Li^{2+}$ ion in $cm^{-1}$?

Similar Questions

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

The value of the work function of a metal $X$ is $3.1 \ eV$. What is its threshold frequency (in $Hz$)? (Given: $h = 6.62 \times 10^{-34} \ J \cdot s$)

An electron revolves in an orbit around a stationary nucleus with atomic number $Z = 5$. The energy required to excite the electron from the $3^{rd}$ orbit to the $4^{th}$ orbit is .......... $eV$.

If the ionization energy of $He^{+}$ is $8.68 \times 10^{-18} \ J$,then the energy of $Be^{3+}$ ion in the second orbit is:- ($Z$ of $Be = 4$)

The nuclear theory of the atom was put forward by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module