Question

In hydrogen atom the kinetic energy of electron is 3.4 electron volt .the distance of the electron from the nucleus is

Answer 1

NOW AS KINETIC ENERGY IS 3.4 eV
so this implies that the electron is in the 1st excited state i.e n=2
$now \: n = 2 \: \\ so \: distance \: from \: the \: nucleus \: = 0.529 \times {10}^{ - 10} \times \frac{ {n}^{2} }{z} \\ = 0.529 \times 4 \times {10}^{ - 10} \\ = 2.116 \times {10}^{ - 10} \: metres$
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Answer 2

Answer:

The distance of the electron from the nucleus is 211.6 pm.

Explanation:

The energy of nth level in hydrogen atom is determined by formula:

$E_n=\frac{-13.6\times Z^2}{n^2} eV$

Let the orbit in which electron is with -3.4 eV is n.

$-3.4 eV=\frac{-13.6\times 1^2}{n^2} eV$

n = 2

Radius of the nth orbit in hydrogen is given by:

$r_n=n^2\times a_o$

$a^o$ = 52.9 pm = Bohr's radius

Radius of the second orbit is:

$r_n=2^2\times 52,9 pm=211.6 pm$

The distance of the electron from the nucleus is 211.6 pm.