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  • A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, then a photon of ene

A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, then a photon of energy 40.8 eV is emitted. The value of n will be:

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Let ground state energy (in eV) be E_{1}

Then, from the given condition,

\mathrm{E}_{2 \mathrm{n}}-\mathrm{E}_1=204 \mathrm{eV}           or           \frac{E_1}{4 n^2}-E_1=204 \mathrm{eV}

\Rightarrow \mathrm{E}_1\left(\frac{1}{4 \mathrm{n}^2}-1\right)=204 \mathrm{eV}                                           (i)

\text { and } E_{2 n}-E_n=40.8 \mathrm{eV}

\Rightarrow \frac{E_1}{4 n^2}-\frac{E_1}{n^2}=E_1\left(-\frac{3}{4 n^2}\right)=40.8 \mathrm{eV}                    (ii)

Form eqs. (i) and (ii), we get

\frac{1-\frac{1}{4 n^2}}{\frac{3}{4 n^2}}=5 \Rightarrow n=2

Posted by

Devendra Khairwa

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