Question

When the hydrogen atom is excited from the ground state to the first excited state, then, (This question has multiple correct options)
A. Its kinetic energy increases by 20eV.
B. Its kinetic energy decreases by 10.2eV.
C. Its potential energy increases by 20.4eV.
D. Its angular momentum increases by $1.05\times {{10}^{-34}}Js$

Answer 1

Hint: According to Bohr’s model of an atom, the electrons can only occupy a set of orbits in which the angular momentum of the electrons is an integral multiple of h-bar($\hbar$). So, the kinetic energy of an electron in orbit depends on the distance from the nucleus and also on which quantized orbit the electron is in. The potential energy of an electron in an orbit depends on the radius of the orbit and the atomic number of the atom in which the excitation takes place.

Formula Used:
From Bohr’s theory, we know that the angular momentum of an electron must be an integral multiple of $\hbar \text{ or }\dfrac{h}{2\pi }$. So, we can write,
$L=n\dfrac{h}{2\pi }$

The potential energy of an electron in orbit is nothing but the electrostatic potential energy of the electron and the nucleus, which can be written as,
$P.E=-k\dfrac{Z{{q}_{p}}{{q}_{e}}}{r}$
Where
Z is the atomic number which means the number of protons in the atom.
k is the proportionality constant whose value is $9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$.
${{\text{q}}_{\text{p}}}\text{ and }{{\text{q}}_{e}}$ are the charges of proton and electron respectively, whose magnitudes are equal.

The kinetic energy of an electron in an orbit is given by,
$K.E=k\dfrac{Z{{q}_{p}}{{q}_{e}}}{2r}$

The radius of the orbit from the nucleus is given by the formula,
$r=\dfrac{{{n}^{2}}}{Z}{{a}_{0}}$ .. equation (5)
Where
Z is the atomic number.
${{a}_{0}}$ is called the Bohr radius, whose value is $0.529\times {{10}^{-10}}m$.

Complete step by step answer:
From Bohr’s theory, we know that the angular momentum of an electron must be an integral multiple of $\hbar \text{ or }\dfrac{h}{2\pi }$. So, we can write,
$L=n\dfrac{h}{2\pi }$ … equation (1)
The necessary centripetal acceleration for an electron revolving around the nucleus is provided by the electrostatic force of attraction between the nucleus and the electron, so we can write,
$\dfrac{m{{v}^{2}}}{r}=k\dfrac{Z{{q}_{p}}{{q}_{e}}}{{{r}^{2}}}$
$m{{v}^{2}}=k\dfrac{Z{{q}_{p}}{{q}_{e}}}{r}$ … equation (2)
Z is the atomic number which means the number of protons in the atom.
k is the proportionality constant whose value is $9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$.
${{\text{q}}_{\text{p}}}\text{ and }{{\text{q}}_{e}}$ are the charges of proton and electron respectively, whose magnitudes are equal.
m is the mass of the electron.
v is the velocity of the electron in orbit.
r is the radius of the orbit.
All of us are familiar with the equation for the kinetic energy of a body, which is $\dfrac{1}{2}m{{v}^{2}}$. So equation (2) can be written in terms of kinetic energy by dividing both sides by 2,
$\dfrac{1}{2}m{{v}^{2}}=k\dfrac{Z{{q}_{p}}{{q}_{e}}}{2r}$
$K.E=k\dfrac{Z{{q}_{p}}{{q}_{e}}}{2r}$ … equation (3)
Coming to the case of potential energy, we are concerned with the potential energy of an electron in an orbit. This is nothing but the electrostatic potential energy of the electron and the nucleus, which can be written as,
$P.E=-k\dfrac{Z{{q}_{p}}{{q}_{e}}}{r}$ … equation (4)
Where
Z is the atomic number which means the number of protons in the atom.
k is the proportionality constant whose value is $9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$.
${{\text{q}}_{\text{p}}}\text{ and }{{\text{q}}_{e}}$ are the charges of proton and electron respectively, whose magnitudes are equal.
So, from equations (1), (3) and (4), we now know the equations for angular momentum, kinetic energy and potential energy respectively.
So, in order to calculate these factors, we need to know the radius of the orbit ‘r’. We can find that using the formula,
$r=\dfrac{{{n}^{2}}}{Z}{{a}_{0}}$ .. equation (5)
Where
Z is the atomic number.
${{a}_{0}}$ is called the Bohr radius, whose value is $0.529\times {{10}^{-10}}m$.
So, for a hydrogen atom, Z=1 and since the electron is excited to the first excited state, the value of n for the ground state should be 1. Therefore, the radius of the orbit is,
$r=\dfrac{{{n}^{2}}}{Z}{{a}_{0}}=1\times {{a}_{0}}$
$\therefore r=0.529\times {{10}^{-10}}m$
So, substituting the value of r in the equation of potential energy, we get,
$P.{{E}_{1}}=-k\dfrac{Z{{q}_{p}}{{q}_{e}}}{r}=-k\dfrac{{{q}_{e}}^{2}}{{{a}_{0}}}$
Substituting the values in the above equation, (${{q}_{e}}=1.6\times {{10}^{-19}}C$), we get,
$P.{{E}_{1}}=-\left( 9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \right)\dfrac{{{\left( 1.6\times {{10}^{-19}}C \right)}^{2}}}{\left( 0.529\times {{10}^{-10}}m \right)}$
$P.{{E}_{1}}=-4.35\times {{10}^{-18}}J$
$\therefore P.{{E}_{1}}=-27.22eV$
For the first excited state, n=2, so $r=2.116\times {{10}^{-10}}m$. So, the P.E in this orbit is,
$P.{{E}_{2}}=-k\dfrac{Z{{q}_{p}}{{q}_{e}}}{r}=-k\dfrac{{{q}_{e}}^{2}}{{{\left( 2 \right)}^{2}}{{a}_{0}}}$
$P.{{E}_{2}}=-\left( 9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \right)\dfrac{{{\left( 1.6\times {{10}^{-19}}C \right)}^{2}}}{\left( 2.11\times {{10}^{-10}}m \right)}$
$P.{{E}_{2}}=-6.80eV$
So, the change in potential energy is given by,
$\Delta P.E=P.{{E}_{2}}-P.{{E}_{1}}$
$\Rightarrow \Delta P.E=\left( -6.80eV \right)-\left( -27.22eV \right)$
$\therefore \Delta P.E=20.4eV$
So, the potential energy increases by $20.4eV$.
The kinetic energy of the electron in the ground state is given by,
$K.{{E}_{1}}=k\dfrac{Z{{q}_{p}}{{q}_{e}}}{2r}=k\dfrac{{{q}_{e}}^{2}}{2{{a}_{0}}}$
Substituting the values in the above equation, (${{q}_{e}}=1.6\times {{10}^{-19}}C$), we get,
$P.{{E}_{1}}=\left( 9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \right)\dfrac{{{\left( 1.6\times {{10}^{-19}}C \right)}^{2}}}{2\times \left( 0.529\times {{10}^{-10}}m \right)}$
$K.{{E}_{1}}=2.17\times {{10}^{-18}}J$
$\therefore K.{{E}_{1}}=13.6eV$
For the first excited state, n=2, so $r=2.116\times {{10}^{-10}}m$. So, the P.E in this orbit is,
$K.{{E}_{2}}=k\dfrac{Z{{q}_{p}}{{q}_{e}}}{2r}=-k\dfrac{{{q}_{e}}^{2}}{2\times {{\left( 2 \right)}^{2}}{{a}_{0}}}$
$K.{{E}_{2}}=\left( 9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \right)\dfrac{{{\left( 1.6\times {{10}^{-19}}C \right)}^{2}}}{2\times \left( 2.11\times {{10}^{-10}}m \right)}$
$K.{{E}_{2}}=3.4eV$
So, the change in kinetic energy is given by,
$\Delta K.E=K.{{E}_{2}}-K.{{E}_{1}}$
$\Rightarrow \Delta K.E=\left( 3.40eV \right)-\left( 13.6eV \right)$
$\therefore \Delta P.E=-10.2eV$
So, the kinetic energy decreases by $10.2eV$.
The change in angular momentum is, (refer equation (1))
$\Delta L={{L}_{2}}-{{L}_{1}}$
$\Delta L=\dfrac{2h}{2\pi }-\dfrac{h}{2\pi }$
$\Delta L=1.05\times {{10}^{-34}}Js$
So, the angular momentum increases by $1.05\times {{10}^{-34}}Js$
So, the answers to the question are options (B), (C) and (D).

Note: We can derive the equation for kinetic energy from the condition for angular momentum from equation (1),
$mvr=\dfrac{nh}{2\pi }$
Where
v is the velocity of the electron in orbit.
r is the radius of the orbit.
m is the mass of the electron.
Divide equation (2) by ‘r’ on both sides and square both sides, then we get,
${{\left( mv \right)}^{2}}={{\left( \dfrac{nh}{2\pi r} \right)}^{2}}$
We know that the kinetic energy of the body is given by the formula,$K.E=\dfrac{1}{2}m{{v}^{2}}$. The above equation can be converted into the equation of kinetic energy if we use both sides of the equation by $2m$. SO, we get,
$\dfrac{1}{2}m{{v}^{2}}=\dfrac{{{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{r}^{2}}m}$
$\therefore K.E=\dfrac{{{n}^{2}}{{h}^{2}}}{8{{\pi }^{2}}{{r}^{2}}m}$
Advantages of Bohr’s theory is that it introduced the idea of quantization in atomic orbits. It specified that the energy and the angular momentum of an electron are quantized.

Disadvantages or limitations to Bohr Theory are:
It is only applicable to single-electron atoms like hydrogen.
It does not explain the splitting of spectral lines in the presence of magnetic and electric fields.
It provides well-defined orbits for the electron to move, and this is not true since an electron can only exist in a space where the probability of finding an electron is high.