An electron in a hydrogen atom first jumps fr | vedclass

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

Question

(D) According to the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$In the first case,the electron jumps from

Vedclass · Accepted Answer

$\frac{20}{7}$

Vedclass · Answer

(D) According to the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} ight]$In the first case,the electron jumps from the third excited state $(n_i = 4)$ to the second excited state $(n_f = 3)$:$\frac{1}{\lambda_1} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{9} - \frac{1}{16} ight] = R \left[ \frac{16 - 9}{144} ight] = \frac{7}{144} R$    .... $(i)$In the second case,the electron jumps from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$:$\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{9 - 4}{36} ight] = \frac{5}{36} R$    .... $(ii)$To find the ratio $\frac{\lambda_1}{\lambda_2}$,we divide the expression for $\frac{1}{\lambda_1}$ by $\frac{1}{\lambda_2}$:$\frac{\lambda_2}{\lambda_1} = \frac{\frac{5}{36} R}{\frac{7}{144} R} = \frac{5}{36} 	imes \frac{144}{7} = \frac{5 	imes 4}{7} = \frac{20}{7}$Therefore,the ratio $\frac{\lambda_1}{\lambda_2} = \frac{20}{7}$.

An electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases is

Similar Questions

In the above figure,$D$ and $E$ respectively represent:

$A$ 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.

An electron is in an excited state in a hydrogen-like atom. It has a total energy of $-3.4 \, eV$. If the kinetic energy of the electron is $E$ and its de-Broglie wavelength is $\lambda$,then:

In a hydrogen atom,if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E,$ the ionization energy of the hydrogen atom is ....... $E$.

Three energy levels of a hydrogen atom and the corresponding wavelengths of the emitted radiation due to different electron transitions are as shown. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module