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 emitting a photon of energy 27.2eV. Alternatively the atom from the same excited state can make a transition to second excited state by emitting a photon of energy 10.20 eV. The value of n and z are given (Ionization energy of hydrogen atom is 13.6eV)

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n=3 and z=6

n=6 and z=3

n=4 and z=8

n=8 and z=4

answer is A.

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Detailed Solution

$\begin{array}{l} ΔE = 13 .6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ 27 .2 = 13 .6 Z^{2} (\frac{1}{2^{2}} - \frac{1}{n^{2}}) \to 1 \\ 10 .2 = 13 .6 Z^{2} (\frac{1}{3^{2}} - \frac{1}{n^{2}}) \to 2 \end{array}$
$\frac{eq (1)}{eq (2)}$ , solving n=6→3
put eq (3) in eq(1) , solving Z=3

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