The wavelength of H_{alpha} line of Balmer se | vedclass

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

Question

(B) The Rydberg formula for the Balmer series $(n_1 = 2)$ is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right)$.For th

Vedclass · Accepted Answer

$X \ \frac{80}{108} \ \mathring{A}$

Vedclass · Answer

(B) The Rydberg formula for the Balmer series $(n_1 = 2)$ is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right)$.For the $H_{\alpha}$ line,$n_2 = 3$: $\frac{1}{X} = RZ^2 \left( \frac{1}{4} - \frac{1}{9} \right) = RZ^2 \left( \frac{5}{36} \right) \implies RZ^2 = \frac{36}{5X}$.For the $H_{\beta}$ line,$n_2 = 4$: $\frac{1}{\lambda_{\beta}} = RZ^2 \left( \frac{1}{4} - \frac{1}{16} \right) = RZ^2 \left( \frac{3}{16} \right)$.Substituting $RZ^2$: $\frac{1}{\lambda_{\beta}} = \left( \frac{36}{5X} \right) \left( \frac{3}{16} \right) = \frac{108}{80X}$.Therefore,$\lambda_{\beta} = X \left( \frac{80}{108} \right) \ \mathring{A}$.

The wavelength of $H_{\alpha}$ line of Balmer series is $X \ \mathring{A}$. What is the wavelength of $H_{\beta}$ line of Balmer series?

Similar Questions

The number of maximum possible different spectral lines obtained in the Balmer series,when an electron makes a transition to the ground state from the $5^{th}$ excited state in a sample of hydrogen atoms,is:

What is the energy associated with the first orbit of $Li^{2+}$ $(R_{H} = 2.18 \times 10^{-18} \ J)$?

The work function of a metal is $4.2 \, eV$. If radiation of $2000 \, \mathring{A}$ falls on the metal,then the kinetic energy of the fastest photoelectron is:

The total energy of an atomic electron is :

Which hydrogen-like species will have the same radius as that of the $1^{st}$ Bohr orbit of a hydrogen atom?

Vedclass Test Series

Exam Paper Generator

Online Exam Module