Question #f42dc

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Answer 1 · 2017-04-13T16:44:45

$"n"_2 = 6$

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, $n_2 = 6$