Question

The wavenumber for the shortest wavelength transition in the Balmer series of atomic hydrogen is:

• A. 27420 $c{m}^{-1}$
• B. 28420 $c{m}^{-1}$
• C. 29420 $c{m}^{-1}$
• D. 12186 $c{m}^{-1}$

Answer 1

The correct option is A 27420 $c{m}^{-1}$

For shortest wavelength ∆E=hcλ,

∆E should be maximum. Thus n1= 2 and n2 = ∞

Therefore, ϑ- = $\frac{1}{\lambda }$ RH [ 122-1∞2]

= 109677 × 142

= 27420 $c{m}^{-1}$