Assertion : Balmer series lies in the visible | vedclass

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

Question

(A) The wavelength for the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $

Vedclass · Accepted Answer

If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

Vedclass · Answer

(A) The wavelength for the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} ight]$,where $n = 3, 4, 5, \dots$For the maximum wavelength $(n=3)$:$\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{5}{36} ight]$$\lambda_{\max} = \frac{36}{5R} \approx 6563 \, \mathring{A}$For the minimum wavelength $(n 	o \infty)$:$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{4} - 0 ight] = \frac{R}{4}$$\lambda_{\min} = \frac{4}{R} \approx 3646 \, \mathring{A}$The range $3646 \, \mathring{A}$ to $6563 \, \mathring{A}$ falls within the visible region of the electromagnetic spectrum. Thus,both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

Assertion : Balmer series lies in the visible region of the electromagnetic spectrum.
Reason : $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $n = 3, 4, 5, \dots$

Similar Questions

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:

The wavelength of the first spectral line in the Balmer series of a hydrogen atom is $6561 \mathring A$. The wavelength of the second spectral line in the Balmer series of a singly-ionized helium atom is

If an electron in a hydrogen atom jumps from the $3^{rd}$ orbit to the $2^{nd}$ orbit,it emits a photon of wavelength $\lambda$. When it jumps from the $4^{th}$ orbit to the $3^{rd}$ orbit,the corresponding wavelength of the photon will be

The difference between the frequencies of the first and second Lyman lines of the hydrogen atom is (where $R$ is the Rydberg constant and $c$ is the speed of light in vacuum).

Using the Rydberg formula,calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion : Balmer series lies in the visible region of the electromagnetic spectrum.Reason : $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $n = 3, 4, 5, \dots$