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.
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
And
r2 = 211.6pm = 52.9n22/Z
Therefore r1/r2 = 1322.5 / 211.6 = n12/n22
or n12/n22 = 6.25
or n1/n2 = 2.5
therefore n2 = 2 , n1 = 5.
Thus the transition is from 5th orbit to 2nd orbit. It belongs Balmer series
Wave number 
for the transition is given by,
1.097 × 107 m–1 (1/22-1/52)
=1.097 x 107m-1 (21/100)
= 2.303 × 106 m–1
Wavelength (λ) associated with the emission transition is given by,
= 0.434 ×10–6 m
λ = 434 nm
the region is visible region
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Answers are very helpfull but pleas do give examples along with examples
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