Question

The number of revolutions done by an electron ‘e’ in one second in the first orbit of hydrogen atom is
a.) $6.57\times {{10}^{15}}$
b.) $6.57\times {{10}^{13}}$
c.) 1000
d.) $6.57\times {{10}^{14}}$

Answer 1

Hint: We know that the electron orbits around the nucleus in a fixed orbit and the velocity of the electron is much higher and due to which we cannot determine both the position and velocity at the same time. This law is known as the Heisenberg uncertainty principle. We all are familiar with this uncertainty principle. In this case we only need to find the revolutions which deal with velocity hence, the uncertainty principle will still be valid but the principle states that we can find any one value with accuracy and we are doing the same here.


Complete step by step solution:
For this question, we should know some basic quantities like,
The speed of electron in 1st orbit of hydrogen is $2.18\times {{10}^{6}}m{{s}^{-1}}$
Radius of orbit $0.0529\times {{10}^{-9}}m$
Now since there is no specific formula to count the number of revolutions, but we can find that by using the formula of time
The formula of time period is given as $T=\dfrac{2\pi n}{v}$
We also know that frequency is defined as $f=\dfrac{1}{T}$
Therefore, frequency will be equal to $f=\dfrac{v}{2\pi r}\text{ }$revolutions per second
Putting the value in above formula we get,
$f=\dfrac{2.18\times {{10}^{6}}}{(2)(3.14)(0.0529\times {{10}^{-9}})}\text{ }$
Solving this we get,
$f=6.57\times {{10}^{15}}$ revolutions per second
Hence, we can say that the electron will make $6.57\times {{10}^{15}}$ revolutions in one second around the first orbit of hydrogen atom
Therefore, we can conclude that option (a) is the correct answer.

Note: Here in this question, we have found the frequency of an electron and as we all know that frequency is defined as how many cycles were repeated in one second. In this case here cycles denote the revolutions as the electron will repeat its revolutions around the orbit and that will become its frequency.