### Question 2.56

NCERT Class XI Chemistry

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.

Note: Q 2.53 to 2.55 are simple enough. So they have been skipped. Please write an email to us if you face any difficulty with any of them.### Solution in Detail

(video solution below this)

$\displaystyle \underline{\underline{\text{Determine Region}}}$

Assuming that hydrogen atom is given

by bohr's theory $\displaystyle r_n = 52.9 \:n^2 \text{ pm}$

$\displaystyle \therefore n = \sqrt{\frac{r_n}{52.9}} \text{, where r}_n\text{ in pm}$

Put r = $\displaystyle 1.3225 \text{ nm} = 1332.5\text{ pm}$

$\displaystyle \therefore n_i = \sqrt{\frac{1332.5}{52.9}} = 5$

Similarly, $\displaystyle n_f = \sqrt{\frac{211.6}{52.9}} = 2$

$\displaystyle n_f = 2 \implies \text{balmer visible region}$

$\displaystyle \underline{\underline{\text{Determine Wavelength}}}$

again, by bohr's theory, for emission

$\displaystyle \overline{\nu} = 1.09677 \times 10^7\bigg(\frac{1}{n_f^2} - \frac{1}{n_i^2}\bigg)\text{ /m}$

$\displaystyle \therefore \overline{\nu} = 1.09677 \times 10^7\bigg(\frac{1}{2^2} - \frac{1}{5^2}\bigg)$

$\displaystyle = 2.303 \times 10^6 \text{ /m}$

$\displaystyle \therefore \lambda = \frac{1}{2.303 \times 10^6} = 434\text{ nm }\underline{Ans}$

### Video Explanation

Please watch this youtube video for a quick explanation of the solution:

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