HardPhysics
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-- 5. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is
A
2430 Å
B
1215 Å
C
1656 Å
D
3286 Å
Explanation
Using 1/λ = RZ²(1/n1² - 1/n2²). For H (Balmer 1st line): 1/6561 = R(1/4 - 1/9) = 5R/36. For He+ (Balmer 2nd line): 1/λ = R(2²)(1/4 - 1/16) = 4R(3/16) = 3R/4. Ratio gives λ = 1215 Å.
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