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

What is the ratio of the total energy of an electron in a hydrogen atom in the first excited state to that in the third excited state?

The total energy of an electron in the ground state of a hydrogen atom is $-13.6 \text{ eV}$. The kinetic energy of an electron in the first excited state is.....$\text{ eV}$

In a hydrogen atom,which of the following transitions will emit a photon of the highest frequency?

The energy of a hydrogen atom in its ground state is $-13.6 \ eV$. The energy of the level corresponding to the quantum number $n = 2$ (first excited state) in the hydrogen atom is......$eV$.

The energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{-13.6}{n^2} \; eV$. The energy of a photon emitted when the electron jumps from the $n = 3$ state to the $n = 2$ state of hydrogen is approximately......$eV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module