Question

Question #f42dc

Answer 1

`"n"_2 = 6`

Explanation:

Given : `vartheta = 1.14 × 10^14 Hz`
`n_1 = 4`

To find `lambda`,
`lambda = c/(vartheta)`

By taking c as `3 × 10^8 ms^-1` and plugging the values we get,

`lambda = 26315 A°`

Now, to find `n_2`, for finding the spectral lines for Hydrogen we use,

`1/(lambda) = R (1/("n"_1^2) - 1/("n"_2^2))`

Use `1/(R) = 911A°`

By plugging the values in the above equation we get,

`"n"_2^2 = 36`

Which is, `n_2 = 6`