The electron in a hydrogen atom first jumps f | vedclass

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

Question

(A) The Rydberg formula for the wavelength of emitted photons is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.For

Vedclass · Accepted Answer

$20/7$

Vedclass · Answer

(A) The Rydberg formula for the wavelength of emitted photons is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the first jump from the third excited state $(n_i = 4)$ to the second excited state $(n_f = 3)$:$\frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16 - 9}{144} ight) = R \left( \frac{7}{144} ight)$.For the second jump from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$:$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9 - 4}{36} ight) = R \left( \frac{5}{36} ight)$.Now,calculating the ratio $\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{R(5/36)}{R(7/144)} = \frac{5}{36} 	imes \frac{144}{7} = \frac{5 	imes 4}{7} = \frac{20}{7}$.

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths,$\lambda_1/\lambda_2$,of the photons emitted in this process is

Similar Questions

The ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(a)$ second to first energy level and $(b)$ highest energy level to the second level is:

Energy levels $A$,$B$ and $C$ of a certain atom correspond to increasing values of energy,i.e.,$E_A < E_B < E_C$. If $\lambda_1$,$\lambda_2$,and $\lambda_3$ are the wavelengths of the photons corresponding to the transitions shown in the figure,then:

$A$ hydrogen atom initially in the ground level absorbs a photon,which excites it to the $n = 4$ level. Determine the wavelength and frequency of the photon.

The energy of a hydrogen atom in its ground state is $-13.6 \, eV$. The energy of the level corresponding to the quantum number $n = 5$ is ...... $eV$.

$A$ difference of $5.4 \ eV$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? (Take $1 \ eV = 1.6 \times 10^{-19} \ J$,$h = 6.625 \times 10^{-34} \ Js$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module