Question

Find the ratio of diameter of electron in 1st Bohr orbit to that in 4th Bohr orbit.

Answer 1

Hint: We will first calculate the radius of orbit an electron in any “nth” orbit of revolution. Once we get the radius of orbit, then the diameter of that particular orbit will be twice the radius. All the formulas will be based on Bohr’s model of an atom.

Complete answer:
For an electron, revolving in an orbit the radius of its orbit is given by the following formula:
$\Rightarrow {{r}_{n}}=0.529\dfrac{{{n}^{2}}}{Z}\overset{0}{\mathop{\text{A}}}\,$
Where,
$n$ is the orbit number in which electron is revolving around the nucleus
$Z$ is the atomic number of the element whose electron, we have taken into consideration
${{r}_{n}}$ is the radius of revolution of the electron in the ${{n}^{th}}$orbit.
Now, it has been given to us that for a certain element we need to find the ratio of diameter in 1st Bohr orbit to that of 4th Bohr orbit.
Therefore, the diameter (say ${{d}_{n}}$ ) could be given by:
$\begin{align}
  & \Rightarrow {{d}_{n}}=2{{r}_{n}} \\
 & \Rightarrow {{d}_{n}}=2\times 0.529\dfrac{{{n}^{2}}}{Z}\overset{0}{\mathop{\text{A}}}\, \\
 & \Rightarrow {{d}_{n}}=1.06\dfrac{{{n}^{2}}}{Z}\overset{0}{\mathop{\text{A}}}\, \\
\end{align}$
Now, the diameter of 1st Bohr orbit will be equal to:
$\Rightarrow {{d}_{1}}=1.06\dfrac{{{1}^{2}}}{Z}\overset{0}{\mathop{\text{A}}}\,$
$\Rightarrow {{d}_{1}}=1.06\dfrac{1}{Z}\overset{0}{\mathop{\text{A}}}\,$ [Let this expression be equation number (1)]
And, diameter of 4th Bohr orbit will be equal to:
 $\Rightarrow {{d}_{4}}=1.06\dfrac{{{4}^{2}}}{Z}\overset{0}{\mathop{\text{A}}}\,$
$\Rightarrow {{d}_{4}}=1.06\dfrac{16}{Z}\overset{0}{\mathop{\text{A}}}\,$ [Let this expression be equation number (2)]
 On dividing equation number (1) by equation number (2), we get:
 $\begin{align}
  & \Rightarrow \dfrac{{{d}_{1}}}{{{d}_{4}}}=\dfrac{1.06\dfrac{1}{Z}\overset{0}{\mathop{\text{A}}}\,}{1.06\dfrac{16}{Z}\overset{0}{\mathop{\text{A}}}\,} \\
 & \Rightarrow \dfrac{{{d}_{1}}}{{{d}_{4}}}=\dfrac{1}{16} \\
\end{align}$
Hence, the ratio of diameter of electron in 1st Bohr orbit to that in 4th Bohr orbit comes out to be $\dfrac{1}{16}$ .

Note:
We saw the use of formula for radius of an orbit in Bohr’s atomic model. It is a very easy and common formula so it should always be remembered thoroughly. Also, these problems are basic problems of ratio and proportionality, so one should always be careful in writing the final answer as it could be mistakenly written as the inverse of what has been asked in the question.