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Question

In the hydrogen atom, the kinetic energy of the electron is $$3.4eV$$. The distance of that electron from the nucleus:


A
2.116˚A
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B
0.529˚A
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C
1.587˚A
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D
21.16˚A
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Solution

The correct option is A $$2.116 \mathring A$$
$$\dfrac{13.6}{n^{2}}=3.4 ev$$
$$n= 2$$
radius $$=0.529\times 2^{2}$$  $$A^{\circ}$$
           $$=2.116A^{\circ}$$
option A is correct

Chemistry

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