Question

(a) Calculate the Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.

Answer 1

Explanation:

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

Formula used for the radius of the $n^{th}$ orbit will be,

$r_n=\frac{n^2\times 52.9}{Z}pm$   (in pm)

where,

$E_n$ = energy of $n^{th}$ orbit

$r_n$ = radius of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) The Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.

Z = 1

$r_3=\frac{3^2\times 52.9}{1} pm$

$r_3=476.1 pm$

476.1 pm is the Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.

b) The energy (in J) of the atom in part (a)

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

$E_3=-13.6\times \frac{1^2}{3^2}eV=1.51 eV$

$1 eV=1.60218\times 10^{-19} Joules$

$1.51 eV=1.51\times 1.60218\times 10^{-19} Joules=2.4210\times 10^{-19} Joules$

$2.4210\times 10^{-19} Joules$ is the energy of n = 3 orbit of a hydrogen atom.

c)  The energy of an Li²⁺ ion when its electron is in the n = 3 orbit.

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

n = 3, Z = 3

$E_3=-13.6\times \frac{3^2}{3^2}eV = -13.6 eV$

$=-13.6eV = -13.6\times 1.60218\times 10^{-19} Joules=2.179\times 10^{-18} Joules$

$2.179\times 10^{-18} Joules$ is the energy of an Li²⁺ ion when its electron is in the n = 3 orbit.

d) The difference in answers is due to change value of Z in the formula which is am atomic number of the element..

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