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A hydrogen like atom with atomic number Z is in higher excited state of quantum number 'n' . This excited state atom can make a transition to the first excited state by successively emitting two photons of energies 10 eV and 68.2eV respectively. Alternatively the atom from the same the same excited state can make a transition to the 2nd excited state by emitting two photons of energies 4.25eV and 5.95eV respectively. Calculate the value of Z


Solution

We have, energy of the electron in the $$n^{th}$$ energy state in any atom is given by,
$$E=\cfrac {-13.6 z^2}{n^2}eV$$

Let the intermediate energy state in the first case be $$x$$. So,
$$\cfrac {z^2}{4}-\cfrac {z^2}{n^2}=\cfrac {10.2+17}{13.6}=2 \longrightarrow (1)$$
Let the intermediate energy state in the $$2^{nd}$$ case be $$y$$. So,
$$\cfrac {z^2}{9}-\cfrac {z^2}{n^2}=\cfrac {4.25+5.95}{13.6}=\cfrac {3}{4} \longrightarrow (2)$$
$$(2)-(1), \cfrac {5}{36}z^2=\cfrac {5}{4}$$
$$\Rightarrow z^2=9,z=3$$

Chemistry

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