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A gas of hydrogen like atoms can absorb radiations of 68 eV. Consequently, the atoms emit radiations of only three different wavelength. All the wave lengths are equal or smaller than that of the absorbed photon. Then choose the correct statement(s)
$(h c = 12400 e V - \overset{0}{A})$

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The lionization energy of the atoms is 489.6 eV.

The atomic number of atoms is 6.

The minimum wavelength of emitted radiation is $28.49 \overset{0}{A}$

Initially the atom was in first excited state

answer is A, B, C, D.

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Detailed Solution

$^{n} C_{2} = 3, n = 3$

So initially atoms was in n = 2.
$E_{3} - E_{2} = 68 e V$
Hence $Z = 6$
$λ_{\min} = \frac{12400}{E_{3} - E_{1}} = 28.49 \overset{0}{A}$

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