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A hydrogen-like atom (atomic number $Z$) is in a higher excited state of quantum number $n$. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17 eV. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV. Find the atomic number $Z.$

A) 2
B) 4
C) 6
D) 3

Energy of first transition $= \pu{17eV}+\pu{10.2eV}= \pu{27.2eV}$ $$27.2=Z^2(\frac{1}{2^2}-\frac{1}{n^2})\to \frac{Z^2}{n^2}=\frac{Z^2}{4}-27.2 \tag{1}$$

Energy of second transition $= \pu{5.95eV}+\pu{4.25eV}= \pu{10.2eV}$ $$10.2=Z^2(\frac{1}{3^2}-\frac{1}{n^2})\to \frac{Z^2}{n^2}=\frac{Z^2}{9}-10.2 \tag{2}$$

Combining $(1)$ and $(2)$: $$\frac{Z^2}{4}-27.2=\frac{Z^2}{9}-10.2 \to Z^2=\frac{17\times36}{5}\to Z=11.06$$

I can’t find where I am going wrong.

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    $\begingroup$ Don't add the energies to start with but take two steps, so try, for example level $m\to n = 10.2$ then $n\to \mathrm{final \;level}= 17$ eV etc where you have to work out m and n and then get to Z. You have two sets of transitions from the same starting level. $\endgroup$ – porphyrin Jul 23 '19 at 15:39
  • $\begingroup$ But why can’t we add the energies? Doesn’t adding it just give the energy of the spectral line? $\endgroup$ – Aditya Jul 23 '19 at 15:55
  • $\begingroup$ Also I tried doing that, but I end up with the same answer. $\endgroup$ – Aditya Jul 23 '19 at 16:09
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    $\begingroup$ According to the formula you are using, energy should be a dimentionless entity. Rydberg formula called so for a good reason. Also, please try to write a concise and short title reflecting the question you are asking. An abstract call for help isn't constructive (that may explain the downvotes and closing as homework despite you have demonstrated your attempt). $\endgroup$ – andselisk Jul 23 '19 at 16:47
  • $\begingroup$ That’s cool, but I tried the way @porphyrin told me, and I still end up with the same equation $\endgroup$ – Aditya Jul 23 '19 at 17:09
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Karsten had the right idea, but the wrong formula.

$$E = \pu{13.6 eV} \cdot Z^2\cdot \left(\dfrac{1}{3^2} - \dfrac{1}{2^2}\right)$$

$$\pu{17.0 eV} = \pu{13.6 eV} \cdot Z^2\cdot 0.138889$$

$$ 8.99999 = Z^2$$

$$ Z = 3.00$$

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  • $\begingroup$ Yup I figured that out. That was a real dumb mistake on my part $\endgroup$ – Aditya Jul 24 '19 at 10:25
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    $\begingroup$ Oops, thanks for fixing my stupid mistake. Should have written it down on paper... $\endgroup$ – Karsten Theis Jul 24 '19 at 16:00
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Edit: I'm leaving this for context, but the correct answer is the one by MaxW.

This problem has four unknowns (n of the excited state, n of the intermediate state going to the first excited state, n of the intermediate state going to the second excited state, and the atomic number Z). You also have four given quantities, the photon energies.

I like the OPs approach of adding the energies - that eliminates two unknowns we don't need for the answer. I would eliminate another unknown (n of the excited state) by asking for the energy to go from the second excited state (n = 3) to the first excited state (n = 2). You get that energy by going from n = 3 to the initial excited state back to n = 2, with an energy of 17.0 eV (see OP's notes).

Now, we just have to plug this into the correct relationship between Z, n(initial) and n(final). The formula the OP used is suspect because there is an energy on the left-hand side, but only unit-less integers on the right hand side. Wikipedia (Bohr model) gives the energy as:

$$E = \pu{13.6 eV} \frac{Z^2}{3^2 - 2^2}\tag{oops, incorrect formula}$$

so

$$Z^2 = E \cdot \frac{3^2 - 2^2}{\pu{13.6 eV}}= \pu{17.0 eV} \frac{5}{\pu{13.6 eV}} = 6.25$$

Finally, we take the square root to get:

$$Z = 2.5$$

This is not an answer that makes sense because we expect Z to be an integer.

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  • $\begingroup$ So what should be the answer, 2 or 3, assuming we approximate the value to the nearest. $\endgroup$ – Aditya Jul 23 '19 at 17:12
  • $\begingroup$ We should either question my answer, or question the question. One thing you could do is just to assume Z = 2, 3, 4, or 6, and calculate n for the excited state. $\endgroup$ – Karsten Theis Jul 23 '19 at 17:15
  • $\begingroup$ You are right though, I made a mistake in my formula. I will try it again, but according to you the answer is 2.5, so my question from the previous comment still stands $\endgroup$ – Aditya Jul 23 '19 at 17:15
  • $\begingroup$ @Aditya I don't have an answer - I posted my calculation to demonstrate my confusion. $\endgroup$ – Karsten Theis Jul 23 '19 at 17:17

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