Question

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

(b) What is the energy (in J) of the atom in part (a)?
(c) What is the energy of an Li²⁺ ion when its electron is in the n = 3 orbit?
(d) Why are the answers to parts (b) and (c) different?

Answer 1

Step-by-step explanation:

`E_n=-13.6* (Z^2)/(n^2)eV`

Formula used for the radius of the
`n^(th)` orbit will be,

`r_n=(n^2* 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=(3^2* 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* (Z^2)/(n^2)eV`

`E_3=-13.6* (1^2)/(3^2)eV=1.51 eV`

`1 eV=1.60218* 10^(-19) Joules`

`1.51 eV=1.51* 1.60218* 10^(-19) Joules=2.4210* 10^(-19) Joules`

`2.4210* 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* (Z^2)/(n^2)eV`

n = 3, Z = 3

`E_3=-13.6* (3^2)/(3^2)eV = -13.6 eV`

`=-13.6eV = -13.6* 1.60218* 10^(-19) Joules=2.179* 10^(-18) Joules`

`2.179* 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* (Z^2)/(n^2)eV`