The wavelength of a spectral line emitted by | vedclass

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

Question

(C) For the Lyman series,the Rydberg formula is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{1^2} - \frac{1}{n_2^2} \right]$.Given $\lambda = \fra

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) For the Lyman series,the Rydberg formula is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{1^2} - \frac{1}{n_2^2} \right]$.Given $\lambda = \frac{16}{15 R}$,so $\frac{1}{\lambda} = \frac{15 R}{16}$.Substituting this into the formula: $\frac{15 R}{16} = R \left[ 1 - \frac{1}{n_2^2} \right]$.Dividing both sides by $R$: $\frac{15}{16} = 1 - \frac{1}{n_2^2}$.Rearranging the terms: $\frac{1}{n_2^2} = 1 - \frac{15}{16} = \frac{1}{16}$.Therefore,$n_2^2 = 16$,which gives $n_2 = 4$.

The wavelength of a spectral line emitted by a hydrogen atom in the Lyman series is $\frac{16}{15 R} \ cm$. What is the value of $n_2$? $(R = \text{Rydberg constant})$

Similar Questions

In which transition is one quantum of energy emitted?

Give names of atomic models and write about Thomson model.

Which of the following electronic transitions in a hydrogen atom will require the highest energy?

In the hydrogen spectrum,if an electron transitions from the $6^{th}$ orbit to the $2^{nd}$ orbit through multiple steps,the number of spectral lines produced is ....

The maximum energy is present in any electron at

Vedclass Test Series

Exam Paper Generator

Online Exam Module