Q5 / 104 · NTA Abhyas (JEE Main) 2020 · Physics · Atomic Physics

If the first excitation energy of a hydrogen-like atom is $27.3 eV$ , then the ionization energy of this atom will be

Correct answer: 36.4

Solution

First excitation energy

$= Z^{2} Rhc (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = Z^{2} Rhc \frac{3}{4}$

$∴ \frac{3}{4} Rhc = 27.3 eV$

$∴ Z^{2} Rhc = \frac{4}{3} 27.3 eV = 36.4 eV$

Hence, the correct answer is 36.4.

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