Thinking Process

Since, the radii of the orbits increase inversely as atomic number $Z$ i.e.,

$$ r \propto \frac{1}{Z} $$

Answer

(c) The atomic number of lithium is 3 , therefore, the radius of $Li^{++}$ion in its ground state, on the basis of Bohr’s model, will be about $\frac{1}{3}$ times to that of Bohr radius.

Therefore, the radius of lithium ion is near $\frac{53}{3} \approx 18 pm$.

Thinking Process

The electron revolves uniformly around nucleus have certain centripetal acceleration associated with it.

Answer

(c) When one decides to work in a frame of reference where the electron is at rest, the given expression is not true as it forms the non-inertial frame of reference.

Thinking Process

The electrostatic force of attraction between electron and nucleus is a central force which provide necessary centripetal force for circular motion of electron.

Answer

(a) The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because of the electrons not being subject to a central force.

Thinking Process

Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\frac{h}{2 \pi}$ where $h$ is the Planck’s constant $(=6.6 \times 10^{-34} J-s)$.

Answer

(a) In the simple Bohr model, only the magnitude of angular momentum is kept equal to some integral multiple of $\frac{h}{2 \pi}$, where, $h$ is Planck’s constant and thus, the Bohr model gives incorrect values of angular momentum.

Thinking Process

The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus.

This happens only because the nuclear force is much stronger than the Coulomb force. The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres.

Answer

(a) In the molecules, nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged.

Thinking Process

The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a 0 . The energy of this state $(n=1), E_1$ is $-13.6 eV$.

Answer

(a) The total energy associated with the two $H$-atoms in the ground state collide in elastically $=2 \times(13.6 eV)=27.2 eV$.

The maximum amount by which their combined kinetic energy is reduced when any one of them goes into first excited state after the inelastic collision.

The total energy associated with the two $H$-atoms after the collision

$$ =(\frac{13.6}{2^{2}})+(13.6)=17.0 eV $$

Therefore, maximum loss of their combined kinetic energy

$$ =27.2-17.0=10.2 eV $$

Thinking Process

The electron of atoms in excited states can fall back to a state of lower energy, emitting a photon in the process.

Answer

(a) A set of atoms in an excited state decays in general to any of the states with lower energy.

Thinking Process

A hydrogen molecule contain two electrons and two protons whereas ionised $H$-molecule consists of an electron and two protons.

Answer

(a, c)

The protons are separated by a small distance of the order of angstrom. In the ground state the electron would not move in circular orbits the electrons, orbit would go around the protons

Answer

$(a, b)$

When beam of free electrons is aiming towards free protons. Then, they scatter but an electron and a proton cannot combine to produce a $H$-atom because of energy conservation and without simultaneously releasing energy in the form of radiation.

Thinking Process

Niel’s Bohr proposed a model for hydrogenic (single electron) atoms in order to explain the line spectra emitted by atoms, as well as the stability of atoms.

Answer

$(a, b)$

The Bohr model for the spectra of a $H$-atom will not be applicable to hydrogen in the molecular form. And also, it will not be applicable as it is for a He-atom.

Thinking Process

The various lines in the atomic spectra are produced when electrons jump from higher energy state to a lower energy state and photons are emitted. These spectral lines are called emission lines.

Answer

$(b, d)$

Balmer series for the $H$-atom can be observed if we measure the frequencies of light emitted due to transitions between higher excited states and the first excited state and as a sequence of frequencies with the higher frequencies getting closely packed.

Thinking Process

When an atom absorbs a photon that has precisely the same energy needed by the electron in a lower energy state to make transitions to a higher energy state, the process is called absorption.

Answer

$(b, d)$

When all the $H$-atoms are in the ground state and radiation of frequency $\frac{(E_2-E_1)}{h}$ falls on it, some of atoms will move to the first excited state and no atoms will make a transition to the $n=3$ state.

Thinking Process

Neil’s Bohr proposed a model for hydrogenic (single electron) atoms in order to explain the line spectra emitted by atoms, as well as the stability of atoms.

Answer

(c, $d)$

The simple Bohr model is not applicable to $He^{4}$ atom because $He^{4}$ has one more electron and electrons are not subject to central forces.

Thinking Process

Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy, say kinetic energy and vice-versa. Einstein gave the famous mass-energy equivalence relation $E=m c^{2}$ where the energy equivalent of mass $m$ is related by the above equation and $c$ is the velocity of light.

Answer

Since, the difference in mass of a nucleus and its constituents, $\Delta M$, is called the mass defect and is given by

$$ \Delta M=[Z m_{p}+(A-Z) m_{n}]-M $$

Also, the binding energy is given by $B=$ mass defect $(\Delta M) \times c^{2}$.

Thus, the mass of a $H$-atom is

$$ m_{p}+m_{e}-\frac{B}{c^{2}}, \text { where } B \approx 13.6 eV \text { is the binding energy. } $$

Thinking Process

Neil’s Bohr proposed a model for hydrogenic (single electron) atoms in order to explain the stability of atoms.

Answer

On removing one electron from $He^{4}$ and $He^{3}$, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass.Also after removing one electron from $He^{4}$ and $He^{3}$ atoms contain one electron and are hydrogen like atoms.

Thinking Process

The accelerated electron produces electric as well as magnetic field hence electromagnetic energy.

Answer

The transition of an electron from a higher energy to a lower energy level can appears in the form of electromagnetic radiation because electrons interact only electromagnetically.

Thinking Process

The electrostatic force of attraction between positively charged nucleus and negatively charged electrons provides necessary centripetal force of revolution. Also, the magnitude of electrostatic force $F \propto q_1 q_2$.

Answer

If proton had a charge $(+4 / 3) e$ and electron a charge $(-3 / 4) e$, then the Bohr formula for the $H$-atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

Thinking Process

Bohr’s postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\frac{h}{2 \pi}$, where $h$ is Planck’s constant $(=6.6 \cdot 10^{-34} J-s)$. Thus, the angular momentum $(L)$ of the orbiting electron is quantised. i.e.,

$$ L=\frac{n h}{2 \pi} $$

Answer

According to Bohr model electrons having different energies belong to different levels having different values of $n$. So, their angular momenta will be different, as

$$ L=\frac{n h}{2 \pi} \text { or } L \propto n $$

Thinking Process

The reduced mass $m$ of two particle system of masses $m_1$ and $m_2$ is given by

$$ \frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2} $$

Answer

The total energy of the electron in the stationary states of the hydrogen atom is given by

$$ E_{n}=-\frac{m e^{4}}{8 n^{2} \varepsilon_0^{2} h^{2}} $$

where signs are as usual and the $m$ that occurs in the Bohr formula is the reduced mass of electron and proton. Also, the total energy of the electron in the ground state of the hydrogen atom is $-13.6 eV$. For $H$-atom reduced mass $m_{e}$. Whereas for positronium, the reduced mass is

$$ m \approx \frac{m_{e}}{2} $$

Hence, the total energy of the electron in the ground state of the positronium atom is

$$ \frac{-13.6 eV}{2}=-6.8 eV $$

Thinking Process

The total energy of the electron in the nth stationary states of the hydrogen. Atom of hydrogen like atom of atomic number $Z$ is given by

$$ E_{n}=z^{2} \frac{-13.6 eV}{n^{2}} $$

Answer

For a $He$-nucleus with charge $2 e$ and electrons of charge $-e$, the energy level in ground state is

$$ -E_{n}=Z^{2} \frac{-13.6 eV}{n^{2}}=2^{2} \frac{-13.6 eV}{1^{2}}=-54.4 eV $$

Thus, the ground state will have two electrons each of energy $E$ and the total ground state energy would be $-(4 \times 13.6) eV=-54.4 eV$.

Thinking Process

The electric current due to revolution of charge is given by $i=\frac{9}{T}=Q \frac{1}{T}=Q \times n$, where $n$ is frequency.

Answer

The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius $(a_{n})$. Let the velocity of electron is $v$.

$\therefore$ Number of revolutions per unit time $=\frac{2 \pi a_0}{v}$

The electric current is given by $i=\frac{q}{t}$, if $q$ charge flows in time $t$. Here, $q=e$

The electric current is given by $i=\frac{2 \pi a_0}{v} e$.

Thinking Process

The problem is based on the explanation of spectrum of hydrogen atom.

Answer

The frequency of any line in a series in the spectrum of hydrogen like atoms corresponding to the transition of electrons from $(n+p)$ level to $n$th level can be expressed as a difference of two terms;

where,

$$ v_{m n}=c R Z^{2} \frac{1}{(n+p)^{2}}-\frac{1}{n^{2}} $$

$m=n+p,(p=1,2,3, \ldots)$ and $R$ is Rydberg constant.

For

$$ \begin{aligned} p & \ll n \\ v_{m n} & =c R Z^{2} \frac{1}{n^{2}} 1+\frac{p}{n}-\frac{1}{n^{2}} \\ v_{m n} & =c R Z^{2} \frac{1}{n^{2}}-\frac{2 p}{n^{3}}-\frac{1}{n^{2}} \end{aligned} $$

[By binomial theorem $(1+x)^{n}=1+n x$ if $.|x|<1]$

$v_{m n}=c R Z^{2} \frac{2 p}{n^{3}} \simeq \frac{2 c R Z^{2}}{n^{3}} p$

Thus, the first few frequencies of light that is emitted when electrons fall to the $n$th level from levels higher than $n$, are approximate harmonic (i.e., in the ratio $1: 2: 3 \ldots$ ) when $n \gg 1$.

Thinking Process

The third line in Balmer series in the spectrum of hydrogen atom is $Hg$.

Answer

$H_{\gamma}$ in Balmer series corresponds to transition $n=5$ to $n=2$. So, the electron in ground state i.e., from $n=1$ must first be placed in state $n=5$.

Energy required for the transition from $n=2$ to $n=5$ is given by

$$ =E_1-E_5=13.6-0.54=13.06 eV $$

Since, angular momentum is conserved,

angular momentum coresponding to $Hg$ photon $=$ change in angular momentum of electron

$$ \begin{aligned} & =L_5-L_2=5 h-2 h=3 h=3 \times 1.06 \times 10^{-34} \\ & =3.18 \times 10^{-34} kg-m^{2} / s \end{aligned} $$

Thinking Process

The reduced mass $m$ of two particle system of masses $m_1$ and $m_2$ is given by

$$ \frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2} $$

Answer

The total energy of the electron in the stationary states of the hydrogen atom is given by

$$ E_{n}=-\frac{m e^{4}}{8 n^{2} \varepsilon_0^{2} h^{2}} $$

where signs are as usual and the $m$ that occurs in the Bohr formula is the reduced mass of electron and proton in hydrogen atom.

By Bohr’s model,

On simplifying,

$$ h v_{i f}=E_{n_{i}}-E_{n_{f}} $$

Since,

$$ v_{i f}=\frac{m e^{4}}{8 \varepsilon_0^{2} h^{3}} \frac{1}{n_f^{2}}-\frac{1}{n_i^{2}} $$

$\lambda \propto \frac{1}{\mu}$

Thus,

$\lambda_{\text {if }} \propto \frac{1}{\mu}$

where $\mu$ is the reduced mass. (here, $\mu$ is used in place of $m$ )

Reduced mass for

$$ \begin{aligned} H=\mu_{H} & =\frac{m_{e}}{1+\frac{m_{e}}{M}} ; m_{e} 1-\frac{m_{e}}{M} \\ D & =\mu_{D} ; m_{e} 1-\frac{m_{e}}{2 M} \\ & =m_{e} 1-\frac{m_{e}}{2 M} 1+\frac{m_{e}}{2 M} \end{aligned} $$

Reduced mass for

If for hydrogen deuterium, the wavelength is $\frac{\lambda_{H}}{\lambda_{D}}$

$$ \begin{aligned} & \frac{\lambda_{D}}{\lambda_{H}}=\frac{\mu_{H}}{\lambda_{D}} \simeq 1+\frac{m_{e}}{2 M} \simeq 1-\frac{1}{2 \times 1840} \\ & \lambda_{D}=\lambda_{H} \times(0.99973) \end{aligned} $$

On substituting the values, we have

Thus, lines are $1217.7 \AA , 1027.7 \AA, 974.04 \AA, 951.143 \AA$.

Thinking Process

The percentage difference in wavelength is given by

$$ 100 \times \frac{\Delta \lambda}{\lambda_{H}}=\frac{\lambda_{D}-\lambda_{H}}{\lambda_{H}} \times 100, \text { where signs are as usual. } $$

Answer

The total energy of the electron in the $n$th states of the hydrogen like atom of atomic number $Z$ is given by

$$ E_{n}=-\frac{\mu Z^{2} e^{4}}{8 \varepsilon_0^{2} h^{2}} \frac{1}{n^{2}} $$

where signs are as usual and the $\mu$ that occurs in the Bohr formula is the reduced mass of electron and proton

Let $\mu_{H}$ be the reduced mass of hydrogen and $\mu_{D}$ that of Deutrium. Then, the frequency of the 1 st Lyman line in hydrogen is $h v_{H}=\frac{\mu_{H} e^{4}}{8 \varepsilon_0^{2} h^{2}} 1-\frac{1}{4}=\frac{\mu_{H} e^{4}}{8 \varepsilon_0^{2} h^{2}} \times \frac{3}{4}$.

Thus, the wavelength of the transition is $\lambda_{H}=\frac{3}{4} \frac{\mu_{H} e^{4}}{8 \varepsilon_0^{2} h^{3} C}$. The wavelength of the transition for the same line in Deutrium is $\lambda_{D}=\frac{3}{4} \frac{\mu_{D} e^{4}}{8 \varepsilon_0^{2} h^{3} c}$.

$\therefore \quad \Delta \lambda=\lambda_{D}-\lambda_{H}$

Hence, the percentage difference is

$$ \begin{aligned} 100 \times \frac{\Delta \lambda}{\lambda_{H}} & =\frac{\lambda_{D}-\lambda_{H}}{\lambda_{H}} \times 100=\frac{\mu_{D}-\mu_{H}}{\mu_{H}} \times 100 \\ & =\frac{\frac{m_{e} M_{D}}{(m_{e}+M_{D})}-\frac{m_{e} M_{H}}{(m_{e}-M_{H})}}{\frac{m_{e} M_{H}}{(m_{e}+M_{H}.}} \times 100 \\ & =\frac{m_{e}+M_{H}}{m_{e}+M_{D}} \frac{M_{D}}{M_{H}}-1 \times 100 \end{aligned} $$

Since, $m_{e} \ll M_{H} \ll M_{D}$

$$ \begin{aligned} \frac{\Delta \lambda}{\lambda_{H}} \times 100 & =\frac{M_{H}}{M_{D}} \times \frac{M_{D}}{M_{H}} \frac{1+\frac{m_{e}}{M_{H}}}{1+\frac{m_{e}}{M_{D}}}-1 \times 100 \\ & =1+\frac{m_{e}}{M_{H}} 1+\frac{m_{e}}{M_{D}}-1 \times 100 \simeq 1+\frac{m_{e}}{M_{H}}-\frac{m_{e}}{M_{D}}-1 \times 100 \\ & \approx m_{e} \frac{1}{M_{H}}-\frac{1}{M_{D}} \times 100 \\ & =9.1 \times 10^{-31} \frac{1}{1.6725 \times 10^{-27}}-\frac{1}{3.3374 \times 10^{-27}} \times 100 \\ & =9.1 \times 10^{-4}[0.5979-0.2996] \times 100 \\ & =2.714 \times 10^{-2} % \end{aligned} $$

Thinking Process

In this problem, expressions are to be derived in two cases namely for a point nucleus in

$H$-atom and for an spherical nucleus of radius $R$

Answer

The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force of revolution.

$$ \frac{m v^{2}}{r_{B}}=-\frac{e^{2}}{r_B^{2}} \cdot \frac{1}{4 \pi \varepsilon_0} $$

By Bohr’s postulates in ground state, we have

$$ m v r=h $$

On solving,

$$ \begin{matrix} \therefore & m \frac{h^{2}}{m^{2} r_B^{2}} \cdot \frac{1}{r_{B}}=+\frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r_B^{2}} \\ \therefore & \frac{h^{2}}{m} \cdot \frac{4 \pi \varepsilon_0}{e^{2}}=r_{B}=0.51 \AA \end{matrix} $$

[This is Bohr’s radius]

The potential energy is given by

$$ \begin{aligned} -\frac{e^{2}}{4 \pi_0} \cdot \frac{1}{r_{B}} & =-27.2 eV ; KE=\frac{m v^{2}}{2} \\ & =\frac{1}{2} m \cdot \frac{h^{2}}{m^{2} r_B^{2}}=\frac{h}{2 m r_B^{2}}=+13.6 eV \end{aligned} $$

Now, for an spherical nucleus of radius $R$,

If $R<r_{B}$, same result.

If $R \gg r_{B}$ the electron moves inside the sphere with radius $r_B^{\prime}(r_B^{\prime}=.$ new Bohr radius).

Charge inside

$$ r_B^{\prime 4}=e \frac{r_B^{\prime 3}}{R^{3}} $$

$$ \therefore \quad \begin{aligned} r_B^{\prime} & =\frac{h^{2}}{m} \frac{4 \pi \varepsilon_0}{e^{2}} \frac{R^{3}}{r_B^{\prime 3}} \\ r_B^{\prime 4} & =(0.51 \AA) \cdot R^{3} \\ & =510(\AA)^{4} \end{aligned} $$

$$ r_B^{\prime 4}=(0.51 \AA) \cdot R^{3} \quad[R=10 \AA] $$

$\therefore \quad r_B^{\prime} \simeq(510)^{1 / 4} \AA<R$

$$ \begin{aligned} KE & =\frac{1}{2} m v^{2}=\frac{m}{2} \cdot \frac{h}{m^{2} r_B^{\prime 2}}=\frac{h}{2 m} \cdot \frac{1}{r_B^{\prime 2}} \\ & =\frac{h^{2}}{2 m r_B^{2}} \cdot \frac{r_B^{2}}{r_B^{\prime 2}}=(13.6 eV) \frac{(0.51)^{2}}{(510)^{1 / 2}}=\frac{3.54}{22.6}=0.16 eV \\ PE & =+\frac{e^{2}}{4 \pi \varepsilon_0} \cdot \frac{r_B^{\prime 2}-3 R^{2}}{2 R^{3}} \\ & =+\frac{e^{2}}{4 \pi \varepsilon_0} \cdot \frac{1}{r_{B}} \cdot \frac{r_B^{\prime}(r_B^{2}-3 R^{2})}{R^{3}}=+(27.2 eV) \frac{0.51(\sqrt{510}-300)}{1000} \\ & =+(27.2 eV) \cdot \frac{-141}{1000}=-3.83 eV \end{aligned} $$

Thinking Process

As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is a single valence electron in $Cr$, the energy states may be thought of as given by the Bohr model.

Answer

The energy of the $n$th state $E_{n}=-Z^{2} R \frac{1}{n^{2}}$ where $R$ is the Rydberg constant and $Z=24$.

The energy released in a transition from 2 to 1 is $\Delta E=Z^{2} R \quad 1-\frac{1}{4}=\frac{3}{4} Z^{2} R$.

The energy required to eject a $n=4$ electron is $E_4=Z^{2} R \frac{1}{16}$.

Thus, the kinetic energy of the Auger electron is

$$ \begin{aligned} KE & =Z^{2} R \frac{3}{4}-\frac{1}{16}=\frac{1}{16} Z^{2} R \\ & =\frac{11}{16} \times 24 \times 24 \times 13.6 eV \\ & =5385.6 eV \end{aligned} $$

Answer

For $m_{p}=10^{-6}$ times, the mass of an electron, the energy associated with it is given by

$$ \begin{aligned} m_{p} c^{2} & =10^{-6} \times \text { electron mass } \times c^{2} \\ & \approx 10^{-6} \times 0.5 MeV \\ & \approx 10^{-6} \times 0.5 \times 1.6 \times 10^{-13} \\ & \approx 0.8 \times 10^{-19} J \end{aligned} $$

The wavelength associated with is given by

where,

$$ \begin{aligned} \frac{\hbar}{m_{p} c} & =\frac{\hbar c}{m_{p} c^{2}}=\frac{10^{-34} \times 3 \times 10^{8}}{0.8 \times 10^{-19}} \\ & \approx 4 \times 10^{-7} m \gg \text { Bohr radius } \\ |\mathbf{F}| & =\frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r^{2}}+\frac{\lambda}{r} \exp (-\lambda r) \end{aligned} $$

$\therefore \quad \lambda \ll \frac{1}{r_{B}}$ i.e., $\lambda r_{B} \ll 1$

Also,

$$ \begin{aligned} & U(r)=-\frac{e^{2}}{4 \pi \varepsilon_0} \cdot \frac{\exp (-\lambda r)}{r} \\ & m v r=\hbar \quad \therefore \quad v=\frac{\hbar}{m r} \end{aligned} $$

$$ \frac{m v^{2}}{r}=\approx \frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r^{2}}+\frac{\lambda}{r} $$

$$ \begin{aligned} & \therefore \quad \frac{\hbar^{2}}{m r^{3}}=\frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r^{2}}+\frac{\lambda}{r} \\ & \therefore \quad \frac{\hbar^{2}}{m}=\frac{e^{2}}{4 \pi \varepsilon_0}[r+\pi r^{2}. \end{aligned} $$

If $\lambda=0$

$$ r=r_{B}=\frac{\hbar}{m} \cdot \frac{4 \pi \varepsilon_0}{e^{2}} $$

$$ \frac{\hbar^{2}}{m}=\frac{e^{2}}{4 \pi \varepsilon_0} \cdot r_{B} $$

Since, $\lambda^{-1} \gg r_{B}$, put $r=r_{B}+\delta$

$$ \begin{matrix} \therefore & r_{B}=r_{B}+\delta+\lambda(r_B^{2}+\delta^{2}+2 \delta r_{B}) ; \text { neglect } \delta^{2} \\ \text { or } & 0=\lambda r_B^{2}+\delta(1+2 \lambda r_{B}) \\ \delta & =\frac{-\lambda r_B^{2}}{1+2 \lambda r_{B}} \approx \lambda r_B^{2}(1-2 \lambda r_{B})=-\lambda r_B^{2} \end{matrix} $$

Since, $\lambda r_{B} \ll 1$

$$ \begin{matrix} \therefore \quad V(r) & =-\frac{e^{2}}{4 \pi \varepsilon_0} \cdot \frac{\exp (-\lambda \delta-\lambda r_{B})}{r_{B}+\delta} \\ \therefore \quad V(r) & =-\frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r_{B}} \quad 1-\frac{\delta}{r_{B}} \cdot(1-\lambda r_{B}) \\ & \cong(-27.2 eV) \text { remains unchanged } \\ KE & =-\frac{1}{2} m v^{2}=\frac{1}{2} m \cdot \frac{h^{2}}{m r^{2}}=\frac{h^{2}}{2(r_{B}+\delta)^{2}}=\frac{h^{2}}{2 r_B^{2}} 1-\frac{2 \delta}{r_{B}} \\ & =(13.6 eV)[1+2 \lambda r_{B}] \\ & =-\frac{e^{2}}{4 \pi \varepsilon_0 r_{B}}+\frac{h^{2}}{2 r_B^{2}}[1+2 \lambda r_{B}] \\ \text { Total energy } & =-27.2+13.6[1+2 \lambda r_{B}] eV \\ & =13.6 \times 2 \lambda r_{B} eV=27.2 \lambda r_{B} eV \end{matrix} $$

Thinking Process

The question offers hypothetical situation in dealing with the total energy of the electron of hydrogen atom.

Answer

Considering the case, when $r \leq R_0=1 \AA$

$$ \begin{aligned} & \text { Let } \varepsilon=2+\delta \\ & \qquad F=\frac{q_1 q_2}{4 \pi \varepsilon_0} \cdot \frac{R_0^{\delta}}{r^{2+\delta}} \end{aligned} $$

where,

$$ \begin{aligned} \frac{q_1 q_2}{4 \pi_0 \varepsilon_0} & =(1.6 \times 10^{-19})^{2} \times 9 \times 10^{9} \\ & =23.04 \times 10^{-29} N m^{2} \end{aligned} $$

The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force.

$$ \begin{aligned} & =\frac{m v^{2}}{r} \text { or } v^{2}=\frac{\Lambda R_0^{\delta}}{m r^{1+\delta}} \\ m v r & =n h \cdot r=\frac{n \hbar}{m v}=\frac{n h}{m} \frac{m}{\Lambda R_0^{\delta}} r^{1 / 2} r^{1 / 2+\delta / 2} \end{aligned} $$

[Applying Bohr’s second postulates]

Solving this for $r$, we get $\quad r_{n}={\frac{n^{2} \hbar^{2}}{m \wedge R_0^{\delta}}}^{\frac{1}{1-\delta}}$

where, $r_{n}$ is radius of $n$th orbit of electron.

For $n=1$ and substituting the values of constant, we get

$$ \begin{aligned} r_1 & =\frac{\hbar^{2}}{m \wedge R_0^{\delta}} \frac{1}{1-\delta} \\ r_1 & =\frac{1.05^{2} \times 10^{-68}}{9.1 \times 10^{-31} \times 2.3 \times 10^{-28} \times 10^{+19}} \\ & =8 \times 10^{-11} \\ & =0.08 nm \end{aligned} $$

$(<0.1 nm)$

This is the radius of orbit of electron in ground state of hydrogen atom. $$ \begin{gathered} v_{n}=\frac{n \hbar}{m r_{n}}=n \hbar \frac{m \wedge R_0^{\delta}}{n^{2} \hbar^{2}} \\ \text { For } n=1, v_1-\frac{\hbar}{m r_1}=1.44 \times 10^{6} m / s \end{gathered} $$

[This is the speed of electron in ground state]

$$ KE=\frac{1}{2} m v_1^{2}-9.43 \times 10^{-19} J=5.9 eV $$

[This is the KE of electron in ground state]

PE till $R_0=-\frac{\Lambda}{R_0}$ [This is the PE of electron in ground state at $r=R_0$ ]

$$ P E \text {from} R_0 \text { to } r=+\Lambda R_0^ \delta $$

[This is the PE of electron in ground state at $R_0$ to $r$ ]

$$ =-\frac{\Lambda R_0^{\delta}}{1+\delta} \frac{1}{r^{1+\delta}}-\frac{1}{R_0^{1+\delta}}=-\frac{\Lambda}{1+\delta} \frac{R_0^{\delta}}{r^{1+\delta}}-\frac{1}{R_0} $$

$$ \begin{aligned} PE & =-\frac{\Lambda}{1+\delta} \frac{R_0^{\delta}}{r^{1+\delta}}-\frac{1}{R_0}+\frac{1+\delta}{R_0} \\ PE & =-\frac{\Lambda}{-0.9} \frac{R_0^{-1.9}}{r^{-0.9}}-\frac{1.9}{R_0} \\ & =\frac{2.3}{0.9} \times 10^{-18}[(0.8)^{0.9}-1.9] J=-17.3 eV \end{aligned} $$

Total energy is $(-17.3+5.9)=-11.4 eV$

This is the required TE of electron in ground state.