Question

The radius of the Bohr orbit in the ground state of a hydrogen atom is 0.5Å. The radius of the orbit of the electron in the third excited state of $H{e^ + }$ will be:
(A) 8Å
(B) 4Å
(C) 0.5Å
(D) 0.25Å

Answer 1

In this problem,we are going to apply the concept of Bohr’s model of an atom and use the expression of radius of nth orbit which is obtained from the postulates of this model.

Formula used:
Radius of nth orbit of the electron in hydrogen like atoms is given as
${r_n} = \dfrac{{{n^2}{r_o}}}{Z}$, where ${r_n}$ is the radius of the nth orbit, n is the value of the orbit, Z is the number of protons or atomic number of the given atom, ${r_o}$ is the Bohr radius.

Complete step by step answer:
We are given to find the radius of the orbit of the electron in the third excited state of
$H{e^ + }$.
Helium atom contains 2 electrons and 2 protons but $H{e^ + }$ contains 2 protons and only one electron. The atoms with only one electron are also known as hydrogen-like atoms.
According to the third postulate of the Bohr’s atomic model, energy of electrons in orbits around the nucleus is quantized.
Electrons jump from one orbit to another by either emitting or absorbing photons. It is given by:
With the help of Bohr’s postulates, it is possible to determine the allowed energies of the atom for different allowed orbits of the electron. The theory thus developed is applicable for hydrogen atoms and ions having one electron.
The radius of nth orbit can be written as
${r_n} = \dfrac{{{n^2}{r_o}}}{Z}$ (for hydrogen like atom/ions)
Where n is orbit number, Z is atomic number, ${r_o} = 0.5Å$ and ${r_o}$ is called the Bohr radius.
We need radius of the third excited state so $n = 3 + 1 = 4$, Helium atom’s atomic number is 2, ${r_o} = 0.5Å$
${r_3} = \dfrac{{{4^2}\left( {0.5} \right)}}{2} \\\ \Rightarrow {r_3} = \dfrac{{16 \times 0.5}}{2} = \dfrac{8}{2} \\\ \Rightarrow {r_3} = 4Å \\\$
Hence the correct choice is Option B, 4Å.

Note: For first excited state of an atom, $n = 2$; therefore for nth excited state of an atom, the orbit will be $\left( {n + 1} \right)$ and for a hydrogen atom the atomic number is always 1 so the radius of the nth orbit will be ${r_n} = {n^2}{r_o}$, where n is the number of the orbit and ${r_o}$ is the Bohr radius. So do not confuse the number of the orbit and the number of the excited state.