Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :

Question

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be : (A) 889.2 nm (B) 642.7 nm (C) 488.9 nm (D) 388.9 nm

Tardigrade Admin · Accepted Answer

(1/660) = R ((1/22) - (1/32)) = (5R/36) ...(1) (1/λ) = R ((1/22) - (1/42)) = (3R/16) ...(2) divide equation (1) with (2) (λ/660) = (5 × 16/6 × 3) λ = (4400/9) = 488.88 = 488.9 nm Option (3)

Q. Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2 $^{n d}$ Balmer line (n = 4 to n = 2) will be :

$889.2 nm$

$642.7 nm$

$488.9 nm$

$388.9 nm$

$\frac{1}{660} = R (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = \frac{5 R}{36} ... (1)$
$\frac{1}{λ} = R (\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}}) = \frac{3 R}{16} ... (2)$
divide equation (1) with (2)
$\frac{λ}{660} = \frac{5 \times 16}{6 \times 3}$
$λ = \frac{4400}{9} = 488.88 = 488.9 nm$
Option (3)

Q. Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :

889.2nm

642.7nm

488.9nm

388.9nm