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How do you calculate the wavelength of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level? Recall that for hydrogen $E_n = -2.18 xx 10^-18 J(1/n^2)$

Md Ariful Islam

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Md Ariful Islam

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Note that you were given only one energy state. If you consider two energy states, from $n = 4$ to $n = 1$, we have:

$E_1 - E_4 = color(blue)(DeltaE)$

$= -2.18xx10^(-18) "J"(1/n_f^2 - 1/n_i^2)$

$= -2.18xx10^(-18) "J"(1/1^2 - 1/4^2)$

$= -2.18xx10^(-18) "J"(15/16)$

$=$ $-color(blue)(2.04xx10^(-18) "J")$

After you obtain the energy, then you can realize that that energy has to correspond exactly to the energy of the photon that came in:

$|DeltaE| = E_"photon" = hnu = (hc)/lambda$

where $h$ is , $c$ is the speed of light, and $lambda$ is the wavelength of the incoming photon. Thus, the wavelength is:

$=> color(blue)(lambda) = (hc)/(E_"photon") = ((6.626xx10^(-34) "J"cdot"s")(2.998xx10^(8) "m/s"))/(2.04xx10^(-18) "J")$

$= 9.720 xx 10^(-8)$ $"m"$

$=$ $color(blue)("97.20 nm")$

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