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

2.116 ˚ A

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0.529 ˚ A

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1.587 ˚ A

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21.16 ˚ A

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Solution

The correct option is A $2.116 ˚ A$
$\frac{13.6}{n^{2}} = 3.4 e v$
$n = 2$
radius $= 0.529 \times 2^{2}$ $A^{\circ}$
$= 2.116 A^{\circ}$
option A is correct

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

The correct option is A 2.116˚A13.6n2=3.4evn=2radius =0.529×22 A∘ =2.116A∘option A is correct

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

The correct option is A $2.116 ˚ A$
$\frac{13.6}{n^{2}} = 3.4 e v$
$n = 2$
radius $= 0.529 \times 2^{2}$ $A^{\circ}$
$= 2.116 A^{\circ}$
option A is correct