How do we work out the frequency of transition for an electronic transition of the hydrogen ATOM, whose electron jumps from $n=2$ to $n=5$?

Md Ariful Islam

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Md Ariful Islam

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And for this process we use the ,

$1/lambda_("vacuum")=R{1/(1/n_1^2-1/n_2^2)}$,.........

......where $R="the Rydberg constant,"$ $1.097 xx 10^7 *m^(−1);$

$n_1=2, n_2=5$........

$1/lambda_("vacuum")=1.097xx10^7*m^-1{1/(1/2^2-1/5^2)}$

$1/lambda_("vacuum")=1.097xx10^7*m^-1{1/(1/4-1/25)}=52238095.2*m^-1$

And thus $lambda=8.90xx10^-8*m$, if I have done my arithmetic correctly, and I would check it!

Now since we deal with the electromagnetic spectrum,

$c=nuxxlambda$, and thus ................

$c/lambda=nu=(3.00xx10^8*m*s^-1)/(8.90xx10^-8*m)=3.371xx10^15*s^-1$

And note that we get an answer in $s^-1$, which is appropriate for a frequency.

So we gots $"wavelength"$, and $"frequency"$, we need to find the energy in $kJ*mol^-1$. We use the old Planck relationship, i.e. $epsilon=hnu$, where $h=6.626xx10^(-34)*J *s=2.23xx10^-18*J$ for a SINGLE PHOTON.

Bu we gots a mole of photons......and thus...

$epsilon_(mol^-1)=2.23xx10^-18*Jxx6.022xx10^23*mol^-1$

$=1.35xx10^6*J*mol^-1=1.35xx10^3*kJ*mol^-1$.

Again you should check my calculations: $"all care taken, but no responsibility admitted."$

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