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Question: The Bohr’s orbit radius for the hydrogen atom (\(n=1\)) is approximately \(0.53A\). The radius for t...

The Bohr’s orbit radius for the hydrogen atom (n=1n=1) is approximately 0.53A0.53A. The radius for the first excited state (n=2n=2) orbit is:
(A). 0.27A0.27A
(B). 1.27A1.27A
(C). 2.12A2.12A
(D). 3.12A3.12A

Explanation

Solution

In the Bohr’s model of atom, only some select energy levels can exist such that the angular momentum of the orbiting electron is an integral multiple ofh2π\dfrac{h}{2\pi }. When an electron gets excited, it absorbs energy and moves from a lower orbit to a higher orbit.

Formulas used:
p=nh2πp=\dfrac{nh}{2\pi }
mv2r=e(ze)4πε0r2\dfrac{m{{v}^{2}}}{r}=\dfrac{e(ze)}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}
mvr=nh2πmvr=\dfrac{nh}{2\pi }
r=0.53n2zr=0.53\dfrac{{{n}^{2}}}{z}

Complete step-by-step solution:
The Bohr’s model of atom has definite circular orbits, also called energy levels, of fixed energy in which electrons revolve around the nucleus. The atom can only have orbits or energy levels in which electrons can have angular momentum as-
p=nh2πp=\dfrac{nh}{2\pi }
Here,
nn is the principal quantum number
pp is the momentum of electron orbiting in thenth{{n}^{th}}orbit
hh is planck's constant

The electrostatic force between the electron and nucleus provides the centripetal force acting on the electron therefore,
mv2r=e(ze)4πε0r2\dfrac{m{{v}^{2}}}{r}=\dfrac{e(ze)}{4\pi {{\varepsilon }_{0}}{{r}^{2}}} - (1)
Here,
ee is the charge on electron
nene is the charge on nucleus
rr is the radius
vv is the tangential velocity
mm is mass of electron
ε0{{\varepsilon }_{0}} is the permeability of free space

Also the angular momentum of the electron is given as-
mvr=nh2πmvr=\dfrac{nh}{2\pi } - (2)

Solving eq (1) and eq (2), we get the relation,
rn2zr\propto \dfrac{{{n}^{2}}}{z} - (3)

Therefore, the expression for radius will be-
r=0.53n2zr=0.53\dfrac{{{n}^{2}}}{z}
Given for hydrogen atom, z=1z=1 first excited state,n=2n=2

In the above equation, we substitute the given values to get,
r=0.53(2)21 r=0.53×4 r=2.12A \begin{aligned} & r=0.53\dfrac{{{(2)}^{2}}}{1} \\\ & \Rightarrow r=0.53\times 4 \\\ & \therefore r=2.12A \\\ \end{aligned}

The radius of the second excited state is 2.12A2.12A. Therefore, the correct option is (C).

Note:
When an electron orbits in some select orbits of Bohr’s model of atom, it never radiates energy and hence it is stable. Bohr’s model of atom is only valid for single electron species. As the radius increases, the force acting on the orbit decreases as the orbit moves farther away from the nucleus.