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The velocity of an electron in a certain Bohr orbit of an H-atom bears the ratio of 1:275 to the velocity of light. The quantum number (n) of this orbit is:A. 3B. 2C. 1D. 4

Last updated date: 24th Jul 2024
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Hint: Recall how the velocity of electrons in the orbits of the Bohr’s model is related to the atomic number of the element and the quantum number of the orbital that specific electron is located in.

Complete step by step solution:
We know that the velocity of an electron as defined by Bohr’s model is the product of the velocity of an electron in the first orbit of a hydrogen atom and the atomic number of the element, this value is divided by the quantum number of the orbital the electron is present in. This formula is denoted as:
${{v}_{e}}=2.19\times {{10}^{6}}m{{s}^{-1}}\times \dfrac{z}{n}$
Where, ${{v}_{e}}$ denotes the velocity of the electron, $z$ denotes the atomic number, and $n$ denotes the quantum number of the orbital. The constant given is the velocity of an electron in the first orbital of the hydrogen atom. In this question, we have to find $n$.
The ratio of the velocity of light to the velocity of the electron is given to us as 275:1. So, it will be shown as:
$\dfrac{{{v}_{e}}}{c}=\dfrac{1}{275}$
Where, $c$is the speed of light i.e. $3\times {{10}^{8}}m/s$. Rearranging the above equation, we get:
${{v}_{e}}=\dfrac{c}{275}$
We will substitute this value, along with the value of $z$ as 1, since the atomic number of hydrogen is 1, in the formula for the velocity of an electron. Then we will rearrange the formula and solve for $n$. We will solve the equation as follows:
\begin{align} & \dfrac{3\times {{10}^{8}}}{275}m{{s}^{-1}}=2.19\times {{10}^{19}}m{{s}^{-1}}\times \dfrac{1}{n} \\ & n=\dfrac{2.19\times {{10}^{6}}\times 275}{3\times {{10}^{8}}} \\ & n=200.75\times {{10}^{-2}} \\ & n=2 \\ \end{align}

Hence, the answer to this question is B ‘2’.

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:
$2\pi r=n\lambda$
$\lambda =\dfrac{h}{mv}$
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}
\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}
\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: While the ratio of the velocities is defined, we will not consider the units but as soon as the value of ${{v}_{e}}$ is determined in terms of the speed of light we have to consider the units of the velocity as $m/s$. Remember that the velocity is directly proportional to the atomic number and inversely proportional to how far away the electron I 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.