Question

Question 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.

Class XI Structure of Atom Page 68

Answer 1

we know, from Bohr's theory ,
radius of nth orbit H like species ,
$r_n=\frac{52.9(n^2)}{Z} pm$
where, n is nth orbit and Z is atomic number .

Given, r1 = 1.3225 nm = 1322.5 pm
now,
$1322.5 pm = \frac{52.9(n_1^2)}{Z}......(1)$
again, r2 = 211.6 pm
$211.6 pm = \frac{52.9(n_2^2)}{Z}........(2)$

now, divide equations (1) and (2)
$\frac{r_1}{r_2}=\frac{1322.5}{211.6}=\frac{n_1^2}{n_2^2}\\\\\frac{n_1^2}{n_2^2}=6.25=(2.5)^2\\\\\frac{n_1}{n_2}=2.5$

it is possible when , n1 = 5 and n2 = 2 so, the transition is from 5th orbit to 2nd orbit .
it belongs to Balmer's series.
now, use formula,
1/λ = 1.09677 × 10^7 ( 1/n1² - 1/n2²)
= 1.09677 × 10^7(1/2² - 1/5²)
= 1.09677 × 10^7 × 21/100
= 2.303 × 10^6 m^-1
λ = 1/2.303 × 10^6 m
= 434 × 10^-9 m = 434 nm

hence, wavelength = 434 nm
it belongs to visible region

Answer 2

Hope this will help you and clear your doubt