An electron in a hydrogen atom drops from energy level n=5 to n=3. What is the energy transition using the Rydberg equation?

Md Ariful Islam

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Md Ariful Islam

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The energy transition will be equal to $1.55 * 10^(-19)"J"$.

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

$1/(lamda) = R * (1/n_("final")^(2) - 1/n_("initial")^(2))$, where

$lamda$ - the wavelength of the emitted photon;
$R$ - Rydberg's constant - $1.0974 * 10^(7)"m"^(-1)$;
$n_("final")$ - the final energy level - in your case equal to 3;
$n_("initial")$ - the initial energy level - in your case equal to 5.

So, you've got all you need to solve for $lamda$, so

$1/(lamda) = 1.0974 * 10^(7)"m"^(-1) * (1/3^2 - 1/5^2)$

$1/(lamda) = 0.07804 * 10^(7)"m"^(-1) => lamda = 1.28 * 10^(-6)"m"$

Since $E = (hc)/(lamda)$, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by $h * c$, where

$h$ - - $6.626 * 10^(-34)"J" * "s"$
$c$ - the speed of light - $"299,792,458 m/s"$

So, the transition energy for your particular transition (which is part of the Paschen Series) is

$E = (6.626 * 10^(-34)"J" * cancel("s") * "299,792,458" cancel("m/s"))/(1.28 * 10^(-6)cancel("m"))$

$E = 1.55 * 10^(-19)"J"$

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