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) 448.9 nm (D) 388.9 nm

Answer 1

Explanation:

take the balmer series formula then use

and solve

Answer 2

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 (C) 448.9 nm

• Using Rydberg equation for Balmer line series in Hydrogen spectrum,

$\frac{1}{L} = R(\frac{1}{{n_f}^{2}} -\frac{1}{{n_i}^{2}} )$ ..........................(1)

where,

L= Wavelength of line spectrum

R = Rydberg constant

nf = final state

ni = initial state

• First Balmer line

nf = 2

ni = 3

L1 = 660 nm

Using (1), we get

$\frac{1}{660} = R (\frac{1}{2^2} - \frac{1}{3^2} ) = R(\frac{1}{4} -\frac{1}{9} )= R(\frac{9-4}{36} )\\\\\frac{1}{660} = R(\frac{5}{36})$......................(2)

• Second Balmer line

nf = 2

ni = 4

L2 = ?

Using (1), we get

$\frac{1}{L2} = R (\frac{1}{2^2} - \frac{1}{4^2} ) \\\\\frac{1}{L2} = R(\frac{1}{4} -\frac{1}{16} )= R(\frac{16-4}{64} )=R(\frac{12}{64}) \\\\\frac{1}{L2} = R(\frac{3}{16})$......................(3)

Dividing (2) by (3), we get

$\frac{\frac{1}{660}}{\frac{1}{L2}} = \frac{R(\frac{5}{36})}{R(\frac{3}{16})}} \\\\\frac{L2}{660} = \frac{5\times 16}{36\times 3} \\\\L2 = \frac{660\times 5\times 16}{36\times 3} \\\\L2 = 488.88\ nm$

Option (C) 448.9 nm is near to the value 488.88 nm. Hence, (C) is the answer.