What is the shortest wavelength for the Pasch | vedclass

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

Question

(C) The Rydberg formula for a hydrogen-like ion is given by: $1 / \lambda = R Z^{2} [ 1 / n_{1}^{2} - 1 / n_{2}^{2} ]$For the Paschen series,the trans

Vedclass · Accepted Answer

$1 / R$

Vedclass · Answer

(C) The Rydberg formula for a hydrogen-like ion is given by: $1 / \lambda = R Z^{2} [ 1 / n_{1}^{2} - 1 / n_{2}^{2} ]$For the Paschen series,the transition ends at $n_{1} = 3$.For the shortest wavelength,the transition starts from $n_{2} = \infty$.For $Li^{2+}$ ion,the atomic number $Z = 3$.Substituting these values: $1 / \lambda = R 	imes 3^{2} [ 1 / 3^{2} - 1 / \infty^{2} ]$$1 / \lambda = R 	imes 9 [ 1 / 9 - 0 ]$$1 / \lambda = R 	imes 9 	imes (1 / 9) = R$Therefore,$\lambda = 1 / R$.

What is the shortest wavelength for the Paschen series of the $Li^{2+}$ ion?

Similar Questions

What is the ratio of the radii of the second orbit for $H$,$He^{+}$,and $Li^{2+}$?

Find out the distance between the $2^{nd}$ and $3^{rd}$ orbit of the $H$-atom $:-$

The difference between the radii of $3^{rd}$ and $4^{th}$ orbits of $Li^{2+}$ is $\Delta R_{1}$. The difference between the radii of $3^{rd}$ and $4^{th}$ orbits of $He^{+}$ is $\Delta R_{2}$. Ratio $\Delta R_{1} : \Delta R_{2}$ is :

The longest wavelength doublet absorption transition is observed at $589 \, nm$ and $589.6 \, nm$. Calculate the frequency of each transition and the energy difference between the two excited states.

The radius of the $2^{nd}$ orbit of $Li^{2+}$ is $x$. The expected radius of the $3^{rd}$ orbit of $Be^{3+}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module