# NCERT 12 Physics Atoms Chapter 12 Exercise

Q.1. Choose the current alternative from the clues given at the end of each statement :
(a) The size of the atom in Thomson’s model is ……. The atomic size in Rutherford’s modal (much greater than/no different from/much less than)
(b) IN the ground state of………., electrons are in stable equilibrium, while in …….. electrons always experience a net force (Thomson’s model/Rutherford’s modal).
(c) A classical atom based on……..is doomed to collapse (Thomson’s model/ Rutherford’s model ).
(d) An atom has a nearly continuous mass distribution in……..but has a highly non uniform mass distribution in………(Thomson’s model/Rutherford’s model).
(e) The positively charged part of the atom possesses most of the mass of atom in….. (Rutherford’s model/both the models).
Sol. (a) No different from.
(b) Thomson’s model, Rutherford’s model.
(c) Rutherford’s model.
(d) Thomson’s model, Rutherford’s model.
(e) Both the models.
Q.2. Suppose you are given a chance to repeat the alpha particle scattering experiment using a thin sheet of solid hydrogen in place of gold foil (hydrogen is a solid at temperature below 14 K). What results do you expect ?
Sol. The basic purpose of scattering experiment is defeated, because solid hydrogen will be a much lighter targe compared to the alpha particle acting as projectile. According to theory of elastic collisions, the target hydrogen will move much faster compared to alpha, after the collision. We cannot determine the size of hydrogen nucleus.
Q.3. What is the shortest wavelength present in the Paschen series of spectral lines ?
Sol. From Rydberg’s formula Q.4. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits from the upper level to the lower level. Q.5. The ground state energy of hydrogen atom is – 13.6 eV. What are the kinetic and potential energies of the electron in this state ? Q.6. A hydrogen atom initially in the ground level absorbs a photon, which excites it to n = 4 level. Determine the wavelength and frequency of photon. Q.7. (a) Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n =1, 2 and 3 levels. (b)Calculate the orbital period in each of these levels.  Q.8. The radius of innermost electron orbit of a hydrogen atom is 5.3What are the radii of n = 2 and n = 3 orbits ? Q.9. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted ?

Sol. In ground state, energy of gaseous hydrogen at room temperature = -13.6 eV. When it is bombarded with 12.5 eV electron beam, the energy becomes -13.6 + 12.5= -1.1 eV. The electron would excitation the electron may jump from n = 3 to n = 2 giving rise to Balmer series. It may also jump form n = 3 to n = 1, giving rise to Lyman series.

Q.10. In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 ⨯ 1011 m with orbital speed 3 ⨯ 104 m/s. (Mass of earth = 6.0 ⨯ 1024 kg). (which help you understand the difference between Thomson’s model and Rutherford’s model better). (a) Is the average angle of deflection of α particles by a thin gold foil predicted by Thomson’s model much less, about the same or much greater than that predicted by Rutherford’s atom model ?
(b)Is the probability of back ward scattering (i.e. scattering of  particles at angles greater than 900)  predicted by Thomson’s model, much less, about the same, or much greater than that predicted by Rutherford’s model ?
(c) Keeping other factors fixed, it is found experimentally, that for a small thickness t, the number of alpha particles scattered at moderate angles is proportional to t. What clues does this linear dependence on t provide ?
(d) In which atom model, is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of alpha particles by a thin foil ?
Sol. (a) About the same. This is because we are talking of average angle of deflection.
(b)Much less, because in Thomson’s model, there is no such massive central core called the nucleus as in Rutherford’s model.
(c)This suggests that scattering is predominantly due to a single collision, because chance of a single collision increases with the number of target atoms which increases linearly with the thickness of the foil.
(d) In Thomson model, positive charge is uniformly distributed in the spherical atom. Therefore, a single collision causes very little deflection. Therefore, average scattering angle can be explained only by considering multiple scattering. Thus it is wrong to ignore multiple scattering in Thomson’s model.
On the contrary, in Rutherford’s model, most of the scattering comes form a single collision. Therefore, multiply scattering may be ignored as a first approximation.
Q.12.The gravitational attraction between electron in a hydrogen atom weaker than the coulomb attraction by a factor of about 10-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting. This much greater than the estimated size of the whole universe !

Q.13. Obtain an expression for the frequency of radiations emitted when a hydrogen atom de-excites from level to level (n – 1). For large n, show that the frequency equals the classical frequency of revolution of the electron in the orbit.
Sol. The frequency v of the emitted radiation when a hydrogen atom  In Bohr’s atom model, Hence for large values of n, classical frequency of revolution of electron in nth orbit is the same as the frequency of radiation emitted when hydrogen atom de-excites from level (n) to level (n – 1).

Q.14. Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size ? Why is an atom not, say, thousand times bigger than its typical size ? The question had greatly puzzled Bohr before he arrived at his famous model of the atom. To stimuluate what he might ell have done before his discovery, let us play with the basic constants of nature and see if we can get a quantity with dimension of length that is roughly equal to known size of an atom (≈〖10〗^(-10) m ).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, m e and c. Determine its numerical value.
(b) You will find that the length obtained (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for something else to get the right atomic size. Construct a quantity with the dimensions of length from h, m e and e and confirm that its numerical value has indeed the correct order of magnitude. this is much smaller than the typical atomic size.
(b) However, when we drop c and use h, m_e and e to construct a quantity, which has dimension of length, the quantity we obtain is Q.15. The total energy of electron in the first excited state of hydrogen atom is about – 3.4 eV
(a) What is kinetic energy of electron in this state ?
(b) What is potential energy of electron in this state ?
(c) Which of the answers above would change if the choice of zero of potential energy is changed ? P.E. = -2 (Kinetic energy)
In this calculation, electric potential and hence potential energy is zero at infinity.
Total energy = P.E. + K.E. = -2 KE + KE = -KE

(a) In the first excited state, total energy = -3.4 eV    ؞ KE = – (-3.4 eV) = + 3.4 eV

(b) P.E. of electron in this first excited state = – 2 KE = – 2 × 3.4 = 3.4 = – 6.8 eV

(c) If zero of potential energy is changed, K. E. does not change and continues to be + 3.4 eV. However, the P.E and total energy of the state would change with the choice of zero of potential energy.

Q.16. If Bohr’s quantisation postulate (angular momentum = n h/2 π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun ?
Sol. Bohr’s quantion postulate is in terms of Planck’s constant (h). But angular momenta associated with planetary motion are = 1070 h (for earth). In terms of Bohr’s quantisation postulate, this will correspond to n = 1070. For such large values of n, the differences in successive energies and angular momenta of the quantised levels are so small, that the levels can be considered as continuous and not discrete.
Q.17. Obtain the first Bohr radius and the ground state energy of a muonic hydrogen atom (i.e. an atom in which a negatively charged muon (μ) of mass about 207 m_e revolves around a proton).  Updated: December 4, 2021 — 3:20 pm