Question

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 1

The radius of the nth orbit of hydrogen-like particles = 0.529n2/Z Å. Consider

r1 = 1.3225 nm or 1322.5 pm = 52.9n12 (say 1)

And

r2 = 211.6pm = 52.9n22/Z (say 2)

which is, r1/r2 = 1322.5 / 211.6 = n12/n22

or n12/n22 = 6.25

or n1/n2 = 2.5

That is,  n2 = 2 , n1 = 5.

Hence, the transition is from the 5th orbit to 2nd orbit. Which belongs to " Balmer series."

Wave number(v bar)  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,

λ= 1/(v bar)

λ = 1/2.303 × 10^6 m

= 0.434 ×10–6 m

λ = 434 nm

Thus, the region is visible region.