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Question

A hydrogen like atom (atomic no $$Z$$) is in a higher excited state of quantum number $$n$$. This excited energies $$10.2\ eV$$ and $$17.00\ eV$$ respectively. Alternatively, the atom from the same excited state can make transition to the second excited state by successively emitting $$2$$ photons of energies $$4.25\ eV$$ and $$5.95\ eV$$ respectively. Determine the values of $$n$$ and $$Z$$. (Ionisation energy of Hydrogen atom $$= 13.6\ Ev)$$.


A
n=5,z=3.
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B
n=6,z=4.
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C
n=6,z=3.
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D
n=6,z=5.
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Solution

The correct option is D $$n = 6, z = 3$$.
For hydrogen like atoms
$$E_n= - 13.6 \times \dfrac{ Z^2}{ n^2}$$ eV / atom
Given, $$E_n – E_2 = 10.2 + 17 = 27.2 eV $$…(i)
$$E_n – E_3 = 4.24 + 5.95 = 10.2 eV$$
∴ $$E_3 – E_2 = 17$$
But,$$ E_3 – E_2 =( - 13.6\times \dfrac{Z^{2}}{9} )– ( - 13.6 \times \dfrac{Z^2}{4})$$
$$= - 13.6\times Z^2 [\dfrac{1}{9} – \dfrac{1}{4}]$$
$$⇒ Z = 3$$
Simlarly,
$$E_n – E_2 = 13.6\times \dfrac{3^2}{n^2} – [- 13.6\times \dfrac{3^2}{2^2}]$$
$$= - 13.6\times [ \dfrac{9}{ n^2 }– \dfrac{9}{ 4}] = - 13.6 \times 9 [4 – \dfrac{n^2 } {4n^2}] …(ii)$$
From eq. (i) and (ii),
$$- 13. 6 \times 9 [ 4 – \dfrac{n^2}{4n^2}] = 27.2$$
⇒$$ n^2 =\dfrac{ 489.6 }{ 13.6} = 36 ⇒ n = 6$$

Chemistry

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