Question

Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.

Answer 1

Explanation:

what is the formula to find the electron in the ground state of hydrogen

Answer 2

Explanation:

Step 1:

Given data in question  

distance between the two charges, $r=0.53 \mathrm{A}=0.53 \times 10^{-10} \mathrm{m}$

Using Coulomb's Law, force,

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$    

$q_{1}=q_{2}=e$

$=9 \times 10^{9} \times \frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(0.53 \times 10^{-10}\right)^{2}}$  

$=9 \times 10^{9} \times \frac{2.56 \times 10^{-38}}{\left(0.53 \times 10^{-10}\right)^{2}}$

$=\frac{23.04 \times 10^{-29}}{0.2809 \times 10^{-20}}$

$=8.2 \times 10^{-8} \mathrm{N}$

Step 2:

Mass of an electron is $\left(M_{e}\right)=9.12 \times 10^{-31} \mathrm{kg}$

Colombian Force provides the necessary centripetal force.

$F_{e}=\frac{m_{e} v^{2}}{r}$  

$8.2 \times 10^{-8}=\frac{9.12 \times 10^{-31} \times v^{2}}{0.53 \times 10^{-10}}$    

$8.2 \times 10^{-8} \times 0.53 \times 10^{-10}=9.12 \times 10^{-31} \times v^{2}$

$4.346 \times 10^{-18}=9.12 \times 10^{-31} \times v^{2}$  

$v^{2}=0.4775 \times 10^{13}$

$v^{2}=4.775 \times 10^{12}$

Step 3:

Squaring on both sides, we get

$v=\sqrt{4.775 \times 10^{12}}$

$v=2.18 \times 10^{6} \mathrm{m} / \mathrm{s}$