If the wavelength of the first line of the Blamer series of hydrogen atom is `6561Å` , the wavelength of the second line of the series should be
a.13122Å b.3280Å c.4860Å d.2187Å

Options

A. `218.7 nm`
B. `328.0 nm`
C. `486.nm`
D. `640.0nm`

Answer

C

Explanation

`bar v_(1) = (1)/(lambda_(1)) = R ((1)/(2^(2)) - (1)/(3^(2))) = (5R)/(36)`
`lambda = (36)/(5R)`
`bar v_(2) = (1)/(lambda_(2)) = R ((1)/(2^(2)) - (1)/(4^(2))) = (3R)/(16)`
`lambda_(2) = (16)/(3R)`
When `(36)/(5R) rarr 656.1 nm`
` (16)/(3R) rArr (656.1 xx 5R xx 16)/(36 xx 3R) = 486 nm`