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Question 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

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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