Comprehension Based MCQ Of Modern Physics, Past Year Questions JEE Advance, Class 12, Physics


17 Questions MCQ Test Class 12 Physics 35 Years JEE Mains &Advance Past year Paper | Comprehension Based MCQ Of Modern Physics, Past Year Questions JEE Advance, Class 12, Physics


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QUESTION: 1

PASSAGE - 1

In a mixture of  H-He+ gas (He+ is singly ionized He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He+ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

Q. The quantum number n of the state finally populated in He+ ions is –

Solution:

For hydrogen like atoms 

i.e., when hydrogen comes to ground state it will release 10.2 eV of energy.

E2 = – 13.6 eV
E3 = – 6.04 eV
E4 = – 3.4eV

Here He+ ion is in the first excited state i.e., possessing energy – 13.6 eV. After receiving energy of + 10.2 eV from excited hydrogen atom on collision, the energy of electron will be (–13.6 + 10.2) eV = –3.4 eV.

This means that the electron will jump to n = 4.

QUESTION: 2

PASSAGE - 1

In a mixture of  H-He+ gas (He+ is singly ionized He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He+ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

Q. The wavelength of light emitted in the visible region by He+ ions after collisions with H atoms is –

Solution:

After collision with hydrogen atom the He+ ion is in its third excited state (n = 4). After that the electron can jump into n = 3.

Since only one option is correct, we need not work out the case of electron jumping from n = 4 to n = 2 .

QUESTION: 3

PASSAGE - 1

In a mixture of  H-He+ gas (He+ is singly ionized He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He+ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

Q. The ratio of the kinetic energy of the n = 2 electron for the H atom to that of He+ ion is –

Solution:

K. E. for hydrogen atom (for n = 2) 

QUESTION: 4

PASSAGE-2

Scientists are working hard to develop nuclear fusion reactor.
Nuclei of heavy hydrogen,  known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D r eaction is  energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of  nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma.
Special techniques are used which confine the plasma for a time t0 before the particles fly away form the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 1014 s/cm3.
It may be helpful to use the following: Boltzmann constant 

Q. In the core of nuclear fusion reactor, the gas becomes plasma because of

Solution:

The collection of  nuclei and electron is known as plasma which is formed due to high temperature inside the reactor core.

QUESTION: 5

PASSAGE-2

Scientists are working hard to develop nuclear fusion reactor.
Nuclei of heavy hydrogen,  known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D r eaction is  energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of  nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma.
Special techniques are used which confine the plasma for a time t0 before the particles fly away form the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 1014 s/cm3.
It may be helpful to use the following: Boltzmann constant 

Q. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10–15 m is in the range

Solution:

Applying conservation of mechanical energy we get Loss of kinetic energy of two deuteron nuclei = Gain in their potential energy.

QUESTION: 6

PASSAGE-2

Scientists are working hard to develop nuclear fusion reactor.
Nuclei of heavy hydrogen,  known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D r eaction is  energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of  nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma.
Special techniques are used which confine the plasma for a time t0 before the particles fly away form the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 1014 s/cm3.
It may be helpful to use the following: Boltzmann constant 

Q. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?

Solution:

For the reading B we get nto > 5 × 1014 which is the Lawson criterion for a reactor to work successfully.

QUESTION: 7

PASSAGE-3

When a particle is restricted to move along x - axis between x  = 0 and x = a, where a  is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a . The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as  Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h = 6.6 × 10–34 J s and e = 1.6 × 10–19 C.

Q. The allowed energy for the particle for a particular value of n  is proportional to

Solution:

The length in which the particle is restricted to move is a. This length is a multiple of λ/2.

⇒ E ∝ a-2 for a particular value of n.

QUESTION: 8

PASSAGE-3

When a particle is restricted to move along x - axis between x  = 0 and x = a, where a  is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a . The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as  Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h = 6.6 × 10–34 J s and e = 1.6 × 10–19 C.

Q. If the mass of the particle is m = 1.0 × 10–30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to

Solution:

For ground state n = 1,
Given m = 1.0 × 10–30 kg, a = 6.6 × 10–9 m

QUESTION: 9

PASSAGE-3

When a particle is restricted to move along x - axis between x  = 0 and x = a, where a  is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a . The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as  Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h = 6.6 × 10–34 J s and e = 1.6 × 10–19 C.

Q. The speed of the particle, that can take discrete values, is proportional to

Solution:

QUESTION: 10

PASSAGE - 4

The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr ’s quantization condition.

Q. A diatomic molecule has moment of inertia I. By Bohr ’s quantization condition its rotational energy in the nth level (n = 0 is not allowed) is

Solution:

According to Bohr ’s quantisation principle 

Rotational kinetic energy 

[∵ L = Iω]

QUESTION: 11

PASSAGE - 4

The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr ’s quantization condition.

Q. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close  Then the moment of inertia of CO molecule about its center of mass is close to

(Take h = 2π × 10 –34 J s)

Solution:

Energy given = change in kinetic energy

QUESTION: 12

PASSAGE - 4

The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr ’s quantization condition.

Q. In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 a.m.u.), where 1 a.m.u.  kg, is close to

Solution:

Centre of mass divides the distance between the point masses in inverse ratio of their masses.

Also the moment of inertia of the system is

 

⇒ d = 1.3 × 10–10 m

QUESTION: 13

PASSAGE - 5

The β -decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory,  a proton (p) and an electron (e) are observed as the decay products of the neutron.
Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, i.e.  around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino  to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 × 106 eV. The kinetic energy carried by the proton is only the recoil energy.

Q. If the anti-neutrino had a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?

Solution:

K should be less than 0.8 × 106 eV as anti-neutrino will have some energy.

QUESTION: 14

PASSAGE - 5

The β -decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory,  a proton (p) and an electron (e) are observed as the decay products of the neutron.
Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, i.e.  around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino  to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 × 106 eV. The kinetic energy carried by the proton is only the recoil energy.

Q. What is the maximum energy of the anti-neutrino?

Solution:

The energy shared between anti-neutrino and electron.
If the energy of electron is almost zero then the maximum energy of anti-neutrino is nearly 0.8 × 106 eV.

QUESTION: 15

PASSAGE - 6

The mass of a nucleus  is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if (m1 + m2) < M. Also two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M' only if (m3 + m4) > M'. The masses of some neutral atoms are given in the table below:

Q. The kinetic energy (in keV) of the alpha particle, when the nucleus  at rest undergoes alpha decay, is

Solution:

Here Δm = [209.982876 – (205.974455 + 4.002603)] × 932 MeV
= 5.422 MeV = 5422 keV

By conservation of linear momentum
Linear momentum of α-particle = linear momentum of lead 

On solving the above two equations we get K.Eα = 5319 keV
(a) is the correct option.

QUESTION: 16

PASSAGE - 6

The mass of a nucleus  is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if (m1 + m2) < M. Also two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M' only if (m3 + m4) > M'. The masses of some neutral atoms are given in the table below:

Q. The correct statement is

Solution:

Only in case of c we have m3 + m4 > M'
(c) is the correct option.
In other cases of fission and in the other case of fusion 

QUESTION: 17

STATEMENT-1 If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change.
STATEMENT-2 When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy.

Solution:

Statement 1 : The wavelength of characteristic X-rays depends on the type of atoms of which the target material is made. It does not depend on the accelerating potential.
Therefore, statement I is true.
Statement 2 : When an electric beam strikes the target in an X-ray tube, part of the kinetic energy is converted into Xray energy. This statement is true. But statement 2 does not explain statement 1.