If the first line of Lyman series has a wavelengths 1215.4 `Å`, the first line of Balmer series is appproximately

Options

A. `4864 Å`
B. `1025.5 Å`
C. `6563 Å`
D. `6400 Å`

Answer

c

Explanation

From hydrogen spectrum, when olectron transist from `n_(2)` orbit to `n_(1)` orbit, the emitted wavelength is given by the formula
`(1)/(lambda_(L)) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
Here, R - Rydberg constant
For lyman series `n_(1) = 1, n_(2) = 2`
For first lyman line,
`(1)/(lambda_(L)) = RZ^(2) ((1)/(1^(2)) - (1)/(2^(2))) implies (1)/(lambda_(L)) = (3)/(4). RZ^(2)`
for Balmer series , `n^(1) = 2, n_(2) = 3,4`
For first balmer line,
`(1)/(lambda_(B)) = RZ^(2) ((1)/(2^(2)) - (1)/(3^(2)) ) = RZ^(2). (5)/(36)`
Dividing Eqs (i) by (ii) we get
`(1//lambda_(L))/(1//lambda_(B)) = (3//4. RZ^(2))/(5//36//RZ^(2)) = (3)/(4) xx (36)/(5) = (27)/(5)`
`:. (lambda_(B))/(lambda_(L)) = (27)/(5)`
Here, `lambda_(L) = 1215.4 Å`
`lambda_(B) = (27)/(5) xx 1215.4 Å`
`= 6563.16 Å prop 6563 Å`