Question

What is the velocity of ${{e}^{-}}$ in the second orbit of the $H$ atom?
A. $2.18\times {{10}^{6}}m/s$
B. $3.27\times {{10}^{6}}m/s$
C. $10.9\times {{10}^{5}}m/s$
D. $21.8\times {{10}^{6}}m/s$

Answer 1

Hint: Recall the Bohr’s model of the atom and what formula it gives for the velocity of the electrons present in the orbits. Think about the relation between the velocity of the electron, the atomic number of the element and the orbit which the electron is in.

Complete step by step answer:
First let us recall the formula that is given by the Bohr’s model for the velocity of the electrons that are orbiting around the nucleus. The formula is defined as:
\[v=2.18\times {{10}^{6}}\times \dfrac{z}{n}\]
Here, $z$ is defined as the atomic number of the element in question and $n$ is the number of the orbital it is placed in.
We know that, for hydrogen, the atomic number is 1. The question asks for the velocity of the electron in the second Bohr orbit of the hydrogen atom. So, the value of $n$ will be 2. We will now put these values in the equation and solve the equation.

\[\begin{align}
  & v=2.18\times {{10}^{6}}\times \dfrac{1}{2} \\
 & v=1.09\times {{10}^{6}}m/s \\
\end{align}\]
We can see that the answer that we get is equal to option C after some modifications. Hence, the correct answer is ‘C'. $10.9\times {{10}^{5}}m/s$’
So, the correct answer is “Option C”.

Additional Information:
The value $2.18\times {{10}^{6}}$ is defined as the velocity of the electron if it was present in the first orbital of the hydrogen atom. We know that an electron shows both waves as well as the particle nature, so we can conclude that the circumference of the circular orbit of the electron should be an integral multiple of its wavelength. So, we get the formula as:
\[2\pi r=n\lambda \]
Where, all the variables take their standard meanings. We know the de Broglie hypothesis defines the de Broglie wavelength as:
\[\lambda =\dfrac{h}{mv}\]

Here, $v$ is the velocity and the other variables take their standard meanings. We will substitute the value for lambda in the earlier equation and rearrange the equation to solve for $v$. So, the equation that we obtain is:
\[\begin{align}
  & 2\pi r=n\times \dfrac{h}{mv} \\
 & v=\dfrac{nh}{2\pi rm} \\
\end{align}\]

Now, we will substitute the known values of all the variables and obtain the value for velocity.
\[\begin{align}
  & n=1 \\
 & h=6.63\times {{10}^{-34}}kg{{m}^{2}}{{s}^{-1}} \\
 & \pi =3.142 \\
 & r=0.529\times {{10}^{-10}}m \\
 & m=9.1\times {{10}^{-31}}kg \\
\end{align}\]

Now, we will substitute these values in the equation and obtain the value for velocity
\[\begin{align}
  & v=\dfrac{1\times 6.63\times {{10}^{-34}}}{2\times 3.142\times 0.529\times {{10}^{-10}}\times 9.1\times {{10}^{-31}}} \\
 & v=2.18\times {{10}^{6}}m/s \\
\end{align}\]

Note: Remember that the velocity is directly proportional to the atomic number and inversely proportional to how far away the electron is from the nucleus i.e. in which orbital it is present. You can find the velocity by proportionality too if you know the velocity of the electron in the first orbit of hydrogen.