Question

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

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:
$v_n={\displaystyle \frac{c\times Z}{137\times n}}$

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 $v_n={\displaystyle \frac{c\times Z}{137\times n}}$, 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
$⟹v={\displaystyle \frac{3\times {10}^8\times 2}{137\times 3}}$
$⟹v={\displaystyle \frac{2}{137}}\times {10}^8$
$\therefore v=1.46\times {10}^6$
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 $\mathrm{\Delta }E=E_i-E_f=R({\displaystyle \frac{1}{n_f^2}}-{\displaystyle \frac{1}{n_i^2}})$ 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.