SELECT * FROM question_mgmt as q WHERE id=1282 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:

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?

Answer:

For an atom = 1/λ = RHZ2 (1/n12-1/n22)

For He+ spectrum Z = 2,  n2=4,  n1= 2

Therefore = 1/λ = RH x 4(1/22-1/42) = 3RH/4

For hydrogen spectrum = 3RH/4,  Z = 1

Therefore = 1/λ = RH X 1(1/n12-1/n22)

Or

RH (1/n12-1/n22) = 3RH/4

Or

1/n12-1/n22  = 3/4

Which can be so for n1=1 & n2 = 2, i.e the transition is from n = 2 to n=1


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