Page No 437: - Chapter 12 Atoms Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 12.14:

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 that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10⁻¹⁰ m).

(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.

(b) You will find that the length obtained in (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. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.

Answer:

(a) Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Mass of an electron, m_e = 9.1 × 10⁻³¹ kg

Speed of light, c = 3 ×10⁸ m/s

Let us take a quantity involving the given quantities as 

Where,

∈₀ = Permittivity of free space

And, 

The numerical value of the taken quantity will be:

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.

(b) Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Mass of an electron, m_e = 9.1 × 10⁻³¹ kg

Planck’s constant, h = 6.63 ×10⁻³⁴ Js

Let us take a quantity involving the given quantities as 

Where,

∈₀ = Permittivity of free space

And, 

The numerical value of the taken quantity will be:

Hence, the value of the quantity taken is of the order of the atomic size.

Question 12.15:

The total energy of an electron in the first excited state of the hydrogen atom is about −3.4 eV.

(a) What is the kinetic energy of the electron in this state?

(b) What is the potential energy of the electron in this state?

(c) Which of the Answers above would change if the choice of the zero of potential energy is changed?

Answer:

(a) Total energy of the electron, E = −3.4 eV

Kinetic energy of the electron is equal to the negative of the total energy.

K = −E

= − (− 3.4) = +3.4 eV

Hence, the kinetic energy of the electron in the given state is +3.4 eV.

(b) Potential energy (U) of the electron is equal to the negative of twice of its kinetic energy.

U = −2 K

= − 2 × 3.4 = − 6.8 eV

Hence, the potential energy of the electron in the given state is − 6.8 eV.

(c) The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic and potential energies, total energy of the system will also change.

Question 12.16:

If Bohr’s quantisation postulate (angular momentum = nh/2π) 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?

Answer:

We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). The angular momentum of the Earth in its orbit is of the order of 10⁷⁰h. This leads to a very high value of quantum levels n of the order of 10⁷⁰. For large values of n, successive energies and angular momenta are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.

Question 12.17:

Obtain the first Bohr’s 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 207m_e orbits around a proton].

Answer:

Mass of a negatively charged muon, 

According to Bohr’s model,

Bohr radius, 

And, energy of a ground state electronic hydrogen atom,

We have the value of the first Bohr orbit, 

Let rμ be the radius of muonic hydrogen atom.

At equilibrium, we can write the relation as:

Hence, the value of the first Bohr radius of a muonic hydrogen atom is

2.56 × 10⁻¹³ m.

We have,

Ee= − 13.6 eV

Take the ratio of these energies as:

Hence, the ground state energy of a muonic hydrogen atom is −2.81 keV.