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

Md Ariful Islam

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Md Ariful Islam

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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:

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

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

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