Question

calculate the wave number of the spectral line of shortest wavelength appearing in the bolmer series of hydrogen spectrum given R=1.09x107/m



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Answer 1

Answer:

According to Balmer formula

v,=1/λ= RH={1/n^2°1-1/n^2°2}

For the Balmer series, ni = 2.

Thus, the expression of wavenumber(ṽ) is given by,

Wave number (ṽ) is inversely proportional to the wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.

For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get:

ṽ= 1.5236 × 106 m–1

Answer 2

Answer:

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