Q.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.

Answer:-

Radius of nth orbit of H-like particles = 0.529n²/Z Å = 52.9n²/Z pm
r₁ = 1.3225 nm = 1322.5 pm = 52.9n₁²
r₂ = 211.6 pm = 52.9n₂²/Z
∴ r₁/r_{2 =} 1322.5 pm/211.6 pm = n₁²/n₂²
⇒ n₁²/n₂² = 6.25
⇒ n₁/n₂ = 2.5
If n₂ = 2, n₁ = 5. Thus the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series.
ṽ = 1.097×10⁷ m^-1 (1/2² – 1/5²) = 1.097×10⁷×21/100 m^-1
λ = 1/ṽ = 100/(1.097×21×10⁷) m = 434×10^-9 m = 434 nm
It lies in visible range.