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Question

The ionization energy of $$H$$ atom is $$x \ kJ$$. The energy required for the electron to jump from $$n = 2$$ to $$n = 3$$ will be:


A
5x kJ
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B
36x5 kJ
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C
5x36 kJ
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D
9x4 kJ
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Solution

The correct option is D $$\dfrac{5x}{36} \ kJ$$
The energy of an electron in the $$n^{th}$$ orbit of hydrogen atom $$=\dfrac{-13.6}{n^2}\ eV$$.
$$I.E.=+13.6\ eV=x\ kJ$$

The energy difference of $$2^{nd}$$ and $$3^{rd}$$ orbit
$$\Delta E=x\left [ \dfrac{1}{4}-\dfrac{1}{9} \right ]\ kJ$$
$$\Delta E=x\times \dfrac{5}{36}\ kJ$$
$$\Delta E= \dfrac{5x}{36} \ kJ$$

Chemistry

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