SELECT * FROM question_mgmt as q WHERE id=1265 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=43 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer:

According to Balmer formula

ṽ=1/λ = RH[1/n12-1/n22]

For the Balmer series, ni = 2.

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

Wave number (ṽ) is inversely proportional to 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


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