If the series limit wavelength of the Lyman series of hydrogen atom is 912 Å, then the series limit wavelength of the Balmer series of the hydrogen atom is

Question

If the series limit wavelength of the Lyman series of hydrogen atom is 912 Å, then the series limit wavelength of the Balmer series of the hydrogen atom is (A) 912 Å (B) 1824 Å (C) 3648 Å (D) 456 Å

Tardigrade Admin · Accepted Answer

For series limit of Balmer series, n2=2, n1=∞ (1/λ )=R[(1/n22) - (1/n12)] =[(1/(2)2)-(1/(∞)2)]=(R/4) ∴ λ =(4/R)=(4/10967800)m =4× 912× 10- 10 m =3648 â«

Q. If the series limit wavelength of the Lyman series of hydrogen atom is $912 Å,$ then the series limit wavelength of the Balmer series of the hydrogen atom is

$912 Å$

$1824 Å$

$3648 Å$

$456 Å$

For series limit of Balmer series,
$n_{2} = 2, n_{1} = \infty$
$\frac{1}{λ} = R [\frac{1}{n _{2}^{2}} - \frac{1}{n _{1}^{2}}]$
$= [\frac{1}{( 2 ) ^{2}} - \frac{1}{( \infty ) ^{2}}] = \frac{R}{4}$
$∴ λ = \frac{4}{R} = \frac{4}{10967800} m$
$= 4 \times 912 \times 1 0^{- 10} m$
$= 3648 Å$

Q. If the series limit wavelength of the Lyman series of hydrogen atom is 912Å, then the series limit wavelength of the Balmer series of the hydrogen atom is

912Å

1824Å

3648Å

456Å