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# If the radius of the first Bohr orbit of the H atom is 0.53 ${A^0}$. The radius of third orbit of $H{e^ + }$ will be:(A) 8.46${ A^0}$(B) 0.705${ A^0}$(C) 1.59${ A^0}$(D) 2.38${ A^0}$

Last updated date: 15th Aug 2024
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Hint: Neil Bohr in 1915 proposed the Bohr model of the atom which came into existence with the modification of Rutherford’s model of an atom present at that time. Rutherford’s model explained that a nucleus is the positively charged section that is surrounded by negatively charged electrons.

We first start by explaining Rutherford's model of an atom. The model proposed by Rutherford assumes that the atom consists of a central nucleus and the electron revolving around it is stable much similar to the sun-planet system.
According to classical electromagnetic theory, we find that the accelerating charged particle emits radiation in the form of electromagnetic waves and eventually the energy of that accelerating electron should decrease. The electron would spiral inward and then fall into the nucleus.
And here comes Niels Bohr. It was he who made modifications in Rutherford’s model of an atom by adding the ideas of the quantum hypothesis.
Bohr said that the electromagnetic theory could not be applied to the processes occurring at the atomic scale. Bohr combined classical and quantum concepts to give his theory in the form of postulates.
Postulates of Bohr’s Model of an Atom are given as follows:
- In an atom, electrons are found to revolve around the positively charged nucleus in a definite circular path called orbits or also known as shells.
- Each orbit or shell has a particular fixed energy and these orbits were known as orbital shells.
The energy levels are represented by the quantum number that ranges from the nucleus side with n = 1 having the lowest energy level to the infinity. The orbits n = 1, 2, 3, 4… are said to be K, L, M, N…. shells respectively.
- The electrons in an atom migrate from a lower energy level to a higher energy level by gaining the fixed required energy and an electron migrates from a higher energy level to lower energy level by losing energy.
After having the brief knowledge of the existence of the Bohr’s orbit let’s proceed with the steps to reach our answer.
We now recall the formula present in the chapter given in the Chemistry NCERT which is stated as follows
$\begin{gathered} {r_n} = \dfrac{{{n^2} \times 0.53{A^0}}}{Z} \\ {r_n} = {\text{radius of }}{{\text{n}}^{th}}{\text{ orbit}} \\ \end{gathered}$
In this above equation the n states the number of orbit and Z represents the atomic number
Let’s verify our formula to find the radius of the second orbit of the hydrogen atom
$\begin{gathered} Z = 1,n = 1 \\ {r_n} = \dfrac{{{1^2} \times 0.53{A^0}}}{1} \\ {r_n} = 0.53{A^0} \\ \end{gathered}$
So we get the value of the radius of the hydrogen atom same as given in the question.
Let’s proceed further with the radius of third orbit of $H{e^ + }$ where the value of the n is 2 and Z is also 2
$\begin{gathered} Z = 2,n = 3 \\ {r_n} = \dfrac{{{3^2} \times 0.53{A^0}}}{2} \\ {r_n} = 2.385{A^0} \\ \end{gathered}$
Therefore, the correct option is (D) 2.38${ A^0}$

Note: Don’t confuse the value of n and Z used in the formula. The n represents the number of orbit and Z represents the atomic number. Students often either replace the value of n and Z or forget to take the value of the Z for the element rather they take it as one for the Hydrogen atom everywhere.