Ncert Exercises & Solutions

2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes the transition from an energy level with n = 4 to an energy level with n = 2?

The ni = 4 to nf = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

E = 2.18 x 10^{- 19} [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}]

Substituting the values in the given expression of E:

E = 2.18 x 10^{- 19} [\frac{1}{4^{2}} - \frac{1}{2^{2}}]

= 2.18 x 10^{- 18} [\frac{1 - 4}{16}]

= 2.18 x 10^{- 18} x (- \frac{3}{16})

E = – (4.0875 ×

10^{- 19}

J)
The negative sign indicates the energy of emission.
Wavelength of light emitted

(λ) = \frac{h c}{E}

(since E = \frac{hc}{λ})

Substituting the values in the given expression of λ:

λ = \frac{(6.626 x 10^{- 34}) (3 x 10^{8})}{4.0875 x 10^{- 19}}

λ = 4.8631 x 10^{- 7} m

= 486.3 x 10^{- 9} m

= 486 mm

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