If {lambda _{max }} is 6563,mathring{A} for t | vedclass

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

Question

(A) For the Balmer series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \do

Vedclass · Accepted Answer

$\lambda = \frac{16}{3R}$

Vedclass · Answer

(A) For the Balmer series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$ For the first line (maximum wavelength),$n = 3$: $\frac{1}{\lambda_{\max}} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$. For the second line,$n = 4$: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16}$. Therefore,$\lambda = \frac{16}{3R}$.

If ${\lambda _{\max }}$ is $6563\,\mathring{A}$ for the Balmer series of a particular atom,then the wavelength of the second line for the Balmer series will be:

Similar Questions

The shortest wavelength in the Balmer series of a hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is:

In an $H$-like atom,when an electron transits from energy state $n=5$ to $n=2$,a photon of wavelength $434 \, nm$ is emitted. What will be the wavelength of the photon emitted when the transition occurs from energy state $n=4$ to $n=2$?

The first wavelength of the Paschen series in hydrogen is $18,800 \, \mathring A$. What is the minimum wavelength of the Paschen series in $\mathring A$?

Which of the following statements is true?

In the hydrogen spectrum,the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Brackett series is $........\,\lambda$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module