Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum

Answers (1)

best_answer

The radius of $n^{th}$ orbit of H-like particles is given by:

$=0.529\times\frac{n^2}{Z} \AA$ Or $=52.9\times\frac{n^2}{Z}\ pm$

Here, starting radius, $r_{1} = 1.3225\ nm = 1322.5\ pm = 52.9n_{1}^2$

Ending radius, $r_{2} = 211.6\ pm = 52.9\left (\frac{n_{2}^2}{Z} \right )$

Therefore, $\frac{r_{1}}{r_{2}} = \frac{1322.5\ pm}{211.6\ pm} = \frac{n_{1}^2}{n_{2}^2}$

$\Rightarrow \frac{n_{1}^2}{n_{2}^2} = 6.25$

$\Rightarrow \frac{n_{1}}{n_{2}} = 2.5$

If $n_{2} = 2$ and $n_{1} = 5$ , then the transition is from $5^{th}$ orbit to $2^{nd}$ orbit.

Therefore, it belongs to the Balmer Series.

Frequency $\nu$ is given by:

$= 1.097\times10^7 m^{-1}\left ( \frac{1}{2^2}-\frac{1}{5^2} \right )$

$= 1.097\times10^7\left (\frac{21}{100} \right )m^{-1}$

Wavelength :

$\lambda = \frac{1}{\nu} = \frac{100}{1.097\times21\times10^7}m = 434\times10^{-9}m = 434\ nm$ .

Therefore, it lies in the visible range.

Posted by

Divya Prakash Singh

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers

anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4

anam-khan

When will CBSE Board exam 2025 date sheet be released for Class 10, 12?

Latest Question