In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is

Question

In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is (A) (5/27) (B) (1/93) (C) (4/9) (D) (3/2)

Tardigrade Admin · Accepted Answer

When an atom comes down from some higher energy level to the first energy level then emitted lines form of Lyman series.

\frac{1}{λ _{L}} = R \frac{1}{1 ^{2}} - \frac{1}{n ^{2}}

where

R

is Rydberg's constant.
When an atom comes from higher energy level to the second level, then Balmer series are obtained.

\frac{1}{λ _{B}} = R \frac{1}{2 ^{2}} - \frac{1}{n ^{2}}

For maximum wavelength

n = 2, \frac{1}{λ _{L}} = R 1 - \frac{1}{( 2 ) ^{2}} = R 1 - \frac{1}{4} = \frac{3 R}{4} \dots . (i)

n = 3, \frac{1}{λ _{B}} = R \frac{1}{( 2 ) ^{2}} - \frac{1}{( 3 ) ^{2}} = R \frac{5}{36} \dots . (ii)

Dividing Eq. (ii) by Eq. (i), we get

\frac{λ _{L}}{λ _{B}} = \frac{5}{27}

Q. In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is

$\frac{5}{27}$

$\frac{1}{93}$

$\frac{4}{9}$

$\frac{3}{2}$

Q. In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is

275​

931​

94​

23​

$\frac{5}{27}$

$\frac{1}{93}$

$\frac{4}{9}$

$\frac{3}{2}$