Question

In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton.How many revolutions per second does the electron make? Hint: What must be true for a force that causes circular motion?Ans: ___ Hz

Answer 1

Answer:

Frequency, $f=6.57\times 10^{15}\ Hz$

Explanation:

It is given that, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton. The electric force acting on the electron is balanced by the centripetal force as :

$\dfrac{kq^2}{r^2}=\dfrac{mv^2}{r}$

v is the speed of electron

$v=\sqrt{\dfrac{ke^2}{mr}}$

$v=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times 0.053\times 10^{-9}}}$

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

The speed of electron is given by :

$v=\dfrac{2\pi r}{t}$

$t=\dfrac{2\pi r}{v}$

$t=\dfrac{2\pi \times 0.053\times 10^{-9}}{2.18\times 10^6}$

$t=1.52\times 10^{-16}\ s$

We know that the number of revolutions per second is called frequency of electron. It is given by :

$f=\dfrac{1}{t}$

$f=\dfrac{1}{1.52\times 10^{-16}}$

$f=6.57\times 10^{15}\ Hz$

So, the total number of revolutions per second make by the electron is $f=6.57\times 10^{15}\ Hz$. Hence, this is required solution.