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.
Given:
Radius, r1 = 1.3225 nm
Radius, r2 = 211.6 pm
The radius of the nth orbit of hydrogen – like particles = 0.529n2/Z Å
Now r1 = 1.3225 nm or 1322.5 pm = 52.9n12 / Z
And
R2 = 211.6pm = 52.9n22/Z
Taking the ratio of r1and r2
So r1/r2=1322.5 / 211.6 = n12/n22
n12/n22 = 6.25
n1/n2 = 2.5
Therefore n2 = 2, n1 = 5
By referring to Balmer Formula:
We know that
Wave Number 
Here n1 = 2 and n2 = 5 and the value of Rh = 109678
Wave Number 
= [109678×21] / 100
= 2303238 / 100
= 23032.38 cm-1
We know that Wave Number = [1] / [Wavelength, λ]
Therefore, Wavelength = [1 / 23032.38]
λ = 4.3417× 10-5 cm
Therefore, the wavelength of the light is 4.34× 10-5 cm
The transition belongs to visible region.