Saved Bookmarks
| 1. |
The wavelength of H_(beta) spectral line of Balmer series is 4860 Å then find out the wavelength of H_(alpha)spectral line of the same series |
|
Answer» Solution :For `H_(beta)` LINE of Balmer series of hydrogen ATOM H `(1)/(lambda_(beta))=R[(1)/(2^(2))-(1)/(4^(2))]=R[(1)/(4)-(1)/(16)]` `(1)/(lambda_(beta))=(3R)/(16).....(1)` `rArr` For `H_(alpha)` line of Balmer series of hydrogen atom `(1)/(lambda_(alpha))=R[(1)/(2^(2))-(1)/(3^(2))]=R[(1)/(4)-(1)/(9)]` `(1)/(alpha)=(5R)/(36)...(2)` `rArr` Taking the ratio of equation (1) and (2) `(lambda_(alpha))/(lambda_(beta))=(3R)/(16)xx(36)/(5R)` `lambda_(alpha)=lambda_(beta)xx(27)/(20)` `=4860xx(27)/(20)` ` :. lambda_(alpha)=6561Å` |
|