What must be the velocity of electrons if their associated wavelength is equal to the longest wavelength line in the Lyman series?

Question

Express your answer using four significant figures.

Should I be using the Balmer equation?
$\nu=(3.2881\times10^15s^(-1))(1/2^2-1/n^2)$

Or should I apply the Rydberg one?
$1/\lambda=R_\infty(1/n_1^{2}-1/n_2^{2})$

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Answer 1 · 2016-11-14T22:28:25

$v=5.987xx10^(3)color(white)(x)"m/s"$

The longest wavelength in the Lyman series corresponds to the

$n=2 -> n=1$

transition and can be calculated using the Rydberg equation

$1/(lamda) = R_(oo) * (1/n_f^2 - 1/n_i^2)$

with

$R_(oo) ~~ 1.097373 * 10^7"m"^(-1)$

$n_f = 1$

$n_i = 2$

Rearrange to solve for $lamda$

$lamda = 1/(R_(oo) * (1 - 1/n_i^2))$

Plug in your values to find

$lamda = 1/(1.097373 * 10^(7)"m"^(-1) * (1 - 1/2^2)) = 1.215 * 10^(-7)"m"$

The wavelength of a moving electron is given by the de Broglie expression:

$lambda=h/(mv)$

$m$ is the mass which is $9.1094xx10^(-31)color(white)(x)"kg"$

$h$ is the Planck Constant which has the value $6.626xx10^(-34)color(white)(x)"J.s"$

Rearranging:

$v=h/(mlambda)$

Putting in the numbers:

$v=(6.626xx10^(-31))/(9.1094xx10^(-31)xx1.215xx10^(-7))color(white)(x)"m/s"$

$v=5.987xx10^(3)color(white)(x)"m/s"$