The wave number of the last line of the Lyman | vedclass

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

Question

(D) The Lyman series in the atomic spectrum of the $H$ atom corresponds to electronic transitions from higher energy levels $(n_2 = 2, 3, 4, \dots, \i

Vedclass · Accepted Answer

$n_2 = \infty$ to $n_1 = 1$

Vedclass · Answer

(D) The Lyman series in the atomic spectrum of the $H$ atom corresponds to electronic transitions from higher energy levels $(n_2 = 2, 3, 4, \dots, \infty)$ to the ground state $(n_1 = 1)$.The wave number $(\bar{
u})$ is given by the Rydberg formula: $\bar{
u} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.The 'last line' of a spectral series corresponds to the transition from the highest possible energy level,which is $n_2 = \infty$,to the series limit level,which is $n_1 = 1$.Therefore,the transition is from $n_2 = \infty$ to $n_1 = 1$.

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

Similar Questions

Consider one $He^{+}$ ion is in excited state $(n = 5)$. Which of the following observations hold true as per the Bohr's model?

The energy required for the ionization of an excited hydrogen atom would be (in $eV$)-

In the Rutherford experiment,if light atoms are used instead of heavy elements,what change is observed in the results?

The values of Planck's constant and the speed of light are $6.63 \times 10^{-34} \ J \ s$ and $3.0 \times 10^8 \ m \ s^{-1}$ respectively. Find the approximate wavelength in nanometers $(nm)$ of a light ray with a frequency of $8 \times 10^{15} \ s^{-1}$.

The work function for a metal is $4 \, eV$. To emit a photo electron of zero velocity from the surface of the metal,the wavelength of incident light should be : ................ $\mathring{A}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module