Taking the Bohr radius as $a_0=53 \mathrm{pm}$, the radius of $\mathrm{Li}^{++}$ion in its ground state, on the basis of Bohr's model, will be about

The binding energy of a H -atom, considering an electron moving around a fixed nuclei (proton), is

$$B=-\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2}(m=\text { electron mass })$$

If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be

$$B=-\frac{M e^4}{8 n^2 \varepsilon_0^2 h^2}(M=\text { proton mass })$$

This last expression is not correct, because

The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because

For the ground state, the electron in the H -atom has an angular momentum $=h$, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

$\mathrm{O}_2$ molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms

Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

A set of atoms in an excited state decays

An ionised H -molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state.

Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H -atom.

The Bohr model for the spectra of a H -atom

The Balmer series for the H -atom can be observed

Let $E_n=\frac{-1 m e^4}{8 \varepsilon_0^2 n^2 h^2}$ be the energy of the $n$th level of H -atom. If all the H -atoms are in the ground state and radiation of frequency $\frac{\left(E_2-E_1\right)}{h}$ falls on it,

The simple Bohr model is not applicable to $\mathrm{He}^4$ atom because

The mass of a H -atom is less than the sum of the masses of a proton and electron. Why is this?

Imagine removing one electron from $\mathrm{He}^4$ and $\mathrm{He}^3$. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why?

When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy?

Would the Bohr formula for the H -atom remain unchanged if proton had a charge $(+4 / 3) e$ and electron a charge $(-3 / 4) e$, where $e=1.6 \times 10^{-19} \mathrm{C}$. Give reasons for your answer.

Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?

Positronium is just like a H -atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?

Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstances, calculate the ground state energy of a He-atom.

Using Bohr model, calculate the electric current created by the electron when the H -atom is in the ground state.

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

What is the minimum energy that must be given to a H -atom in ground state so that it can emit an $H_\gamma$ line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such $H_\gamma$ photon?

The first four spectral in the Lyman series of a H-atom are $\lambda=1218\mathop A\limits^o$, $1028\mathop A\limits^o$ and $951.4\mathop A\limits^o$. If instead of Hydrogen, we consider deuterium, calculate the shift in the wavelength of these lines.

Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ${ }^1 \mathrm{H}$ and ${ }^2 \mathrm{H}$. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass.

Such a system is equivalent to a single particle with a reduced mass $\alpha$, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here $\propto=m_e M /\left(m_e+M\right)$, where $M$ is the nuclear mass and $m_e$ is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in ${ }^1 \mathrm{H}$ and ${ }^2 \mathrm{H}$. (mass of ${ }^1 \mathrm{H}$ nucleus is $1.6725 \times 10^{-27} \mathrm{~kg}$, mass of ${ }^2 \mathrm{H}$ nucleus is $3.3374 \times 10^{-27}$ kg , Mass of electron $\left.=9.109 \times 10^{-31} \mathrm{~kg}.\right)$

If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H -atom when

(i) $R=0.1 \mathop A\limits^o$ and(ii) $R=10 \mathop A\limits^o$.

In the Auger process, an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom (This is called an Auger, electron). Assuming the nucleus to be massive, calculate the kinetic energy of an $n=4$ Auger electron emitted by Chromium by absorbing the energy from a $n=2$ to $n=1$ transition.

The inverse square law in electrostatic is $|\mathbf{F}|=\frac{e^2}{\left(4 \pi \varepsilon_0\right) r^2}$ for the force between an electron and a proton. The $\left(\frac{1}{r}\right)$ dependence of $|\mathbf{F}|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$, force would be modified to $|\mathbf{F}|=\frac{e^2}{\left(4 \pi \varepsilon_0\right) r^2}\left[\frac{1}{r^2}+\frac{\lambda}{r}\right] \cdot \exp (-\lambda r)$ where $\lambda=\frac{m_p c}{\hbar}$ and $\hbar=\frac{h}{2 \pi}$. Estimate the change in the ground state energy of a H -atom if $m_p$ were $10^{-6}$ times the mass of an electron.

The Bohr model for the H-atom relies on the Coulomb's law of electrostatics. Coulomb's law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb's law between two opposite charge $+q_1,-q_2$ is modified to

$$\begin{aligned} |\mathbf{F}| & =\frac{q_1 q_2}{\left(4 \pi \varepsilon_0\right)} \frac{1}{r^2}, r \geq R_0 \\ & =\frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{1}{R_0^2}\left(\frac{R_0}{r}\right)^{\varepsilon}, r \leq R_0 \end{aligned}$$

Calculate in such a case, the ground state energy of a H -atom, if $E=0.1$, $R_0=1 \mathop A\limits^o$.