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 = $\frac{\mathrm{h}}{2\mathrm{\pi }}$, 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,

O₂ 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,

(a) the electron would not move in circular orbits
(b) the energy would be ${\left(2\right)}^4$ times that of a H-atom
(c) the electron's orbit would go around the protons
(d) the molecule will soon decay in a proton and a H-atom

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,

(a) Because of energy conservation
(b) Without simultaneously releasing energy in the form of radiation
(c) Because of momentum conservation
(d) Because of angular momentum conservation

The Bohr model for the spectra of a H-atom:

(a) will not be applicable to hydrogen in the molecular form
(b) will not be applicable as it is for a He-atom
(c) is valid only at room temperature
(d) predicts continuous as well as discrete spectral lines

The Balmer series for the H-atom can be observed:

(a) if we measure the frequencies of light emitted when an excited atom falls to the ground state
(b) if we measure the frequencies of light emitted due to transitions between excited states and the first excited state
(c) in any transition in a H-atom
(d) as a sequence of frequencies with the higher frequencies getting closely packed

Let $E_n=\frac{-1}{8{\mathrm{\epsilon }}_0^2}\frac{m{e}^4}{n^2{h}^2}$ be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency $\frac{(E_2-E_1)}{h}$ falls on it,

(a) it will not be absorbed at all
(b) some of the atoms will move to the first excited state
(c) all atoms will be excited to the n = 2 state
(d) no atoms will make a transition to the n = 3 state

The simple Bohr model is not applicable to He⁴ atom because

(a) He⁴ is an inert gas
(b) He⁴ has neutrons in the nucleus
(c) He⁴ has one more electron
(d) electrons are not subject to central forces

The total energy of an electron in the $n^{th}$ stationary orbit of the hydrogen atom can be obtained by:

In Rutherford’s nuclear model of the atom, the nucleus (radius about 10^-15 m) is analogous to the sun about which the electron move in orbit (radius ≈ 10^-10 m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, then:

(The radius of the earth's orbit is about 1.5 x 10¹¹m. The radius of the sun is taken as 7 x 10⁸m.)

In a Geiger-Marsden experiment, what is the distance of the closest approach to the nucleus of a 7.7 MeV $\alpha$-particle before it comes momentarily to rest and reverses its direction?

It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. The velocity of the electron in a hydrogen atom is:

$$\begin{gathered} 1. 3.2\times {10}^6 \mathrm{m}/\mathrm{s} \\ 2. 2.2\times {10}^6 \mathrm{m}/\mathrm{s} \\ 3. 4.2\times {10}^6 \mathrm{m}/\mathrm{s} \\ 4. 1.2\times {10}^6 \mathrm{m}/\mathrm{s} \end{gathered}$$

According to the classical electromagnetic theory, the initial frequency of the light emitted by the electron revolving around a proton in the hydrogen atom is:

(The velocity of the electron moving around a proton in a hydrogen atom is 2.2×${10}^{-6}$ m/s)

$$\begin{gathered} 1. 7.6\times {10}^{13} \mathrm{Hz} \\ 2. 4.7\times {10}^{15} \mathrm{Hz} \\ 3. 6.6\times {10}^{15} \mathrm{Hz} \\ 4. 5.2\times {10}^{13} \mathrm{Hz} \end{gathered}$$

A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom. The quantum number of the orbit of the satellite is:

$$\begin{gathered} 1. 2.0\times {10}^{43} \\ 2. 4.7\times {10}^{45} \\ 3. 3.0\times {10}^{43} \\ 4. 5.3\times {10}^{45} \end{gathered}$$

The wavelength of the first spectral line of the Lyman series of the hydrogen spectrum is: