If the wavelength of the first line of the Balmer series of hydrogen is `6561 Å`, the wavelngth of the second line of the series should be

Options

A. `1312Å`
B. `3280Å`
C. `4860Å`
D. `2187Å`

Answer

C

Explanation

For Balmer series, `n_(1)=2, n_(2)=3` for `1^("st")` line and `n_(2)=4` for second line.
`therefore(lambda_(1))/(lambda_(2))=(((1)/(2^(2))-(1)/(4^(2))))/(((1)/(2^(2))-(1)/(3^(2))))=(3//16)/(5//36)=(3)/(16)xx(36)/(5)=(27)/(20)`
or `lambda_(2)=(20)/(27) lambda_(1)=(20)/(27)xx6561=4860Å`