Question

Calculate the wavelength for the emission transition

Answer 1

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