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A unelectronic species is in some excited state 'A' and on absorbing a photon of energy 'x' eV gets promoted to a new state 'B'. On do-expiation back to the ground state a total of 10 different wavelengths were emitted in which seven have energy greater than 'x' eV What will he the ionization energy?

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$25 x eV$

$\frac{225}{16} x eV$

$\frac{25}{16} x eV$

$25 \times 9 x \in V$

answer is B.

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Detailed Solution

The ionization energy, is :

$Δ E = 13.6 \times z^{2} (\frac{1}{3^{2}} - \frac{1}{5^{2}}) = x$

$13.6 \times \frac{z^{2}}{1^{2}} = E_{1}$

$\frac{E_{1}}{x} = \frac{9 \times 25}{16} \Rightarrow E_{1} = \frac{225}{16} x eV$

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