Get Answers to all your Questions

header-bg qa

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