The minimum and maximum values of wavelength in the Lyman series of a H atom are, respectively
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a
364.3 nm and 653.4 nm
b
91.2 nm and 121.5 nm
c
41.2 nm and 102.6 nm
d
9.12 nm and 121.5 nm
answer is B.
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Detailed Solution
H-atom, Z = 1Lyman series, n1 = 1Minimum value of wavelength (Maximum value of wave number), n2 = ∞v¯=1λ=RH112-1∞2λ=1RH=91.2 nmMaximum value of wavelength (Minimum value of wave number), n2 = n1 + 1 = 1 + 1 = 2v¯=1λ=RH112-122=3RH4λ=43RH =4/3 (91.2) = 121.5 nm