The wavelength of the spectral line obtained | vedclass

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

Question

(D) Let $n_1$ be the lower energy level and $n_2$ be the higher energy level.Given: $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$.Adding the two equations: $2n_

Vedclass · Accepted Answer

$1.14 	imes 10^{-6} \ cm$

Vedclass · Answer

(D) Let $n_1$ be the lower energy level and $n_2$ be the higher energy level.Given: $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$.Adding the two equations: $2n_2 = 6 \implies n_2 = 3$.Subtracting the two equations: $2n_1 = 2 \implies n_1 = 1$.Using Rydberg's formula: $\frac{1}{\lambda} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For $Li^{2+}$, $Z = 3$. Substituting the values: $\frac{1}{\lambda} = R_H (3)^2 \left[ \frac{1}{1^2} - \frac{1}{3^2} ight]$.$\frac{1}{\lambda} = 9 R_H \left[ 1 - \frac{1}{9} ight] = 9 R_H \left( \frac{8}{9} ight) = 8 R_H$.$\lambda = \frac{1}{8 R_H}$.Taking $R_H \approx 1.097 	imes 10^5 \ cm^{-1} \approx 1.1 	imes 10^5 \ cm^{-1}$.$\lambda = \frac{1}{8 	imes 1.1 	imes 10^5} \ cm = \frac{1}{8.8 	imes 10^5} \ cm \approx 0.1136 	imes 10^{-5} \ cm = 1.136 	imes 10^{-6} \ cm$.Thus, $\lambda \approx 1.14 	imes 10^{-6} \ cm$.

The wavelength of the spectral line obtained in the spectrum of $Li^{2+}$ ion, when the transition takes place between two levels whose sum is $4$ and difference is $2$, is

Similar Questions

The ionization energy of a hydrogen atom is $13.6 \ eV$. The energy required to ionize a hydrogen atom in its second excited state is ........... $eV$.

Among the following,which transition in the hydrogen spectrum would have the same wavelength as the Balmer transition,$n=4$ to $n=2$ in the $He^{+}$ spectrum?

$A$ certain electronic transition from an excited state to the ground state of the $H$ atom in one or more steps gives rise to four lines in the ultraviolet region of the spectrum. How many lines does this transition produce in the infrared region of the spectrum?

Heat treatment of muscular pain involves radiation of wavelength of about $900 \ nm$. Which spectral line of $H$ atom is suitable for this? Given: Rydberg constant $R_{H}=10^5 \ cm^{-1}, h=6.6 \times 10^{-34} \ J \ s, c=3 \times 10^8 \ m / s$.

The difference in wavelength of first and second spectral lines of Lyman series in $Li^{2+}$ ion spectrum is ................ $\mathring{A}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module