The minimum and maximum values of wavelength in the Lyman series of a H atom are, respectively:

364.3 n m a n d 653.4 n m

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91.2 n m a n d 121.5 n m

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41.2 n m a n d 102.6 n m

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9.12 n m a n d 121.5 n m

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Solution

The correct option is B $91.2 n m a n d 121.5 n m$
For lyman series,
Maximum wavelength implies minimum transition energy which is for $n_{2} = 2 \to n_{1} = 1, E = 10.4 e V$ .
Minimum wavelength implies maximum transition energy which is for $n_{2} = \infty \to n_{1} = 1, E = 13.6 e V$ .
Therefore,
$λ_{m a x} = h c / E_{m i n} = 121 n m, λ_{m i n} = h c / E_{m a x} = 91 n m$ .

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The minimum and maximum values of wavelength in the Lyman series of a H atom are, respectively:

The correct option is B 91.2nmand121.5nmFor lyman series,Maximum wavelength implies minimum transition energy which is for n2=2→n1=1,E=10.4eV.Minimum wavelength implies maximum transition energy which is for n2=∞→n1=1,E=13.6eV.Therefore,λmax=hc/Emin=121nm,λmin=hc/Emax=91nm.