Number of visible lines in Balmer's series

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

Question

(B) The Balmer series corresponds to transitions of electrons in a hydrogen atom to the $n = 2$ energy level from higher energy levels $(n = 3, 4, 5,

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The Balmer series corresponds to transitions of electrons in a hydrogen atom to the $n = 2$ energy level from higher energy levels $(n = 3, 4, 5, 6, \dots)$.Among these,the transitions from $n = 3, 4, 5,$ and $6$ to $n = 2$ result in wavelengths that fall within the visible spectrum.Specifically,the transitions are:$1$. $n = 3 	o n = 2$ ($H_{\alpha}$ line,red)$2$. $n = 4 	o n = 2$ ($H_{\beta}$ line,blue-green)$3$. $n = 5 	o n = 2$ ($H_{\gamma}$ line,blue-violet)$4$. $n = 6 	o n = 2$ ($H_{\delta}$ line,violet)Transitions from $n > 6$ to $n = 2$ result in wavelengths shorter than $364.6 \ nm$,which lie in the ultraviolet region.Therefore,there are exactly $4$ visible lines in the Balmer series.

Similar Questions

If ${\lambda _{{\text{max}}}}$ is $6563 \text{ Å}$, then the wavelength of the second line of the Balmer series will be:

The ratio of longest wavelengths of the spectral lines in the Lyman and Balmer series of hydrogen spectrum is

If $R$ is the Rydberg constant in $cm^{-1}$, then the hydrogen atom does not emit any radiation of wavelength in the range of

The ratio of the maximum wavelengths of the Lyman and Balmer series in the hydrogen spectrum is ........

The ratio of minimum wavelengths of Lyman and Balmer series will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module