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Q.

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, a photon of energy 40.8 eV is emitted. The value of n will be

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a

1

b

2

c

4

d

3

answer is B.

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Detailed Solution

Let ground state energy (in eV) be $E_{1}$
Then from the given condition

$E_{2 n} - E_{1} = 204 eV \Rightarrow \frac{E_{1}}{4 n^{2}} - E_{1} = 204 eV \Rightarrow E_{1} (\frac{1}{4 n^{2}} - 1) = 204 eV . . . . . . . (i) and$

$E_{2 n} - E_{n} = 40.8 eV \Rightarrow \frac{E_{1}}{4 n^{2}} - \frac{E_{1}}{n^{2}} = E_{1} (- \frac{3}{4 n^{2}}) = 40.8 eV . . . . . . . . (ii) From equations (i) and (ii), \frac{1 - \frac{1}{4 n^{2}}}{\frac{3}{4 n^{2}}} = 5 \Rightarrow n = 2$

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