Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the wave number of 1st  line of Balmer series of H-atom is ‘x’ then the wave number of 1st line of lyman series of the He+ ion will be<div c

If the wave number of 1st  line of Balmer series of H-atom is ‘x’ then the wave number of 1st line of lyman series of the He+ ion will be

Option: 1

\frac{36 x}{5}


Option: 2

\frac{12x}{5}


Option: 3

\frac{108x}{5}


Option: 4

x


Answers (1)

best_answer

Lyman Series spectrum -

 \frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

Where n_{1}=1\ and\ n_{2}=2, 3, 4.....

This lies in the Ultraviolet region

 

Balmer Series Spectrum -

\frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

Where n_{1}=2\ and\ n_{2}=3, 4, 5, 6 .....

It lies in the visible region.

 

\\ \overline{v} = RZ^{2}\left ( \frac{1}{n^{2}_{1}}-\frac{1}{n_{2}^{2}} \right )=R.1\left ( \frac{1}{2^{2}}- \frac{1}{3^{2}} \right )=\frac{5R}{36}\\ x=\frac{5R}{36}\\ \\R=\frac{36x}{5}

\overline{v}_{1} =R\times 2^{2}\left ( 1-\frac{1}{2^{2}} \right )=\frac{36x}{5} \times 3=\frac{108x}{5}

Therefore, Option(3) is correct.

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 exam date this month; updates on reduced syllabus, registration, number of attempts
anam-khan

AP BArch Admission 2024: Registration begins today at apsche.ap.gov.in; eligibility criteria

Latest Question