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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 quantium state n, a photon of energy

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 quantium state n, a photon of energy 40.8 eV is emitted. Then vale of Z is:

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

Let ground state energy (in eV) be \mathrm{E_1}. Then, from the given condition

\mathrm{\begin{aligned} & E_{2 n}-E_1=204 \mathrm{eV} \\ & \text { or } \frac{E_1}{4 n^2}-E_1=204 \mathrm{eV} \\ & \text { or } E_1\left(\frac{1}{4 n^2}-1\right) \quad=204 \mathrm{eV} \end{aligned}}        [1]

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

\mathrm{\begin{aligned} \frac{E_1}{4 n^2}-\frac{E_1}{n^2} & =40.8 \mathrm{eV} \\ E_1\left(\frac{-3}{4 n^2}\right) & =40.8 \mathrm{eV} \end{aligned}}               [2]

From Eq. Number (1) and (2),

\mathrm{\begin{aligned} \frac{1-\frac{1}{4 n^2}}{\frac{3}{4 n^2}} & =5 \\ 1 & =\frac{1}{4 n^2}+\frac{15}{4 n^2} \\ \frac{4}{n^2} & =1 \end{aligned}}

or n = 2
From Eq. Number (2),

\mathrm{\begin{aligned} E_1 & =\frac{-4}{3} n^2(40.8) \mathrm{eV} \\ & =\frac{4}{3}(2)^2(40.8) \mathrm{eV} \\ \text { or } \quad E_1 & =-217.6 \mathrm{eV} \\ E_1 & =-(13.6) Z^2 \\ \therefore \quad Z^2 & =\frac{E_1}{-13.6}=16 \\ \therefore \quad Z & =4 \end{aligned}}

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Kshitij

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