Question

The wave number of a line of Lyman series of H-spectrum is 8R/9 (R=Rybergb constant). The transition is

Answer 1

Answer-$3 \Rightarrow 2$

Explanation:

For lyman series

$\frac{1}{\lambda} =R(\frac{1}{(1)^2}-\frac{1}{(n_2)^2} )$

$\frac{8R}{9}=R(\frac{1}{1}-\frac{1}{(n_2)^2} )$

$\frac{8}{9}=1-\frac{1}{(n_2)^2}$

$\frac{1}{(n_2)^2}=1-\frac{8}{9}$

$\frac{1}{(n_2)^2}=\frac{9-8}{9} =\frac{1}{9}$

${(n_2)^2}=9$

$n_2=\sqrt{9}=3$

$\therefore n_2=3$

Hence, a transition is from $3\Rightarrow 2$

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Answer 2

Answer:-

The transition is 3⇔1

Explanation:

Given:-

H-spectrum is 8R/9 (R=Rybergb constant).

In physics and chemistry, the Lyman series is a hydrogen spectral series of transitions and leads to the ultraviolet emission lines of a hydrogen atom, while an electron from n ≥ 2 to n = 1 (where n is the principal quantum number) is the lowest. energy levels. electron.

Wave number of spectral line in emission spectrum of hydrogen,

vˉ=RH(n121−n221) …(i)

Given, vˉ=98RH

On putting the value of vˉ in Eq. (i), we get-

98RH=RH(n121−n221)

98=(1)21−n221

98−1=n221

31=n21∴n2=3

Hence, The Transition is  n2=3 to n1=1

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