Question

Consider $3^{rd}$ orbit of $He^{+}$ (Helium), using non-relativistic approach, the speed of electrons in this orbit will be (given $K=9\times 10^{9}$ constant, $Z=2$ and $h=6.6\times 10^{-34}Js$)
\[\begin{align}
  & A.0.73\times {{10}^{6}}m/s \\
 & B.3.0\times {{10}^{8}}m/s \\
 & C.2.92\times {{10}^{6}}m/s \\
 & D.1.46\times {{10}^{6}}m/s \\
\end{align}\]

Answer 1

Hint: From the Bohr -Rutherford model of an atom, we know that the nucleus of the atom is present in the centre of the atom and the electrons revolve around the nucleus. The Bohr- Rutherford model of the atom, we have the velocity of the electron in the nth orbit. Using this formula, we can solve this question.

Formula used:
$v_{n}=\dfrac{c\times Z}{137\times n}$

Complete step-by-step answer:
We know from the Bohr -Rutherford model of an atom that velocity $v_{n}$ of the electrons at the n-th orbit is given as $v_{n}=\dfrac{c\times Z}{137\times n}$, where $c$ is the speed of the light, $Z$ is the atomic number of the atom and $n$ is the number of the orbit.
Here, it is given that $Z=2$ and $n=3$, then substituting the values, we have
$\implies v=\dfrac{3\times 10^{8}\times 2}{137\times 3}$
$\implies v=\dfrac{2}{137}\times 10^{8}$
$\therefore v=1.46\times 10^{6}$
Hence the correct answer is option \[D.1.46\times {{10}^{6}}m/s\]

So, the correct answer is “Option D”.

Additional Information: Similarly, for hydrogen atom which is a simple atom, to which the Bohr -Rutherford model of an atom can be applied. We know that the Bohr -Rutherford model of an atom, gives the relationship of the emission spectrum due to a hydrogen atom.
The spectral lines are due to the excitation of the electrons from on energy level of the hydrogen atom to the other, and it is given as $\Delta E=E_{i}-E_{f}=R\left(\dfrac{1}{n_{f}^{2}}- \dfrac{1}{n_{i}^{2}}\right)$ where $E_{i}$ is the initial energy of the hydrogen, when the electron is in $n_{i}$ state and $E_{f}$ is the final energy of the hydrogen, when the electron is in $n_{f}$ state. And $R$ is the Rydberg constant
Since the transition of the electrons occurs in the visible region, then we can say that $E=h\nu$, where $h$ is the Planck’s constant and $\nu$ is the frequency of the light.

Note: The above is the simplest way to find the velocity of any given atom using the Bohr -Rutherford model of an atom. Also, not that here, we are not using the values of $K$ or $h$ which are given in the question. Be careful with the calculations.