JEE Advanced Physics Q&A – Miscellaneous Mixed Concepts Problems

Q 1 :

An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use? [2016]

64
90
108
120

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(3)

$Using,$ $\frac{A}{A_{0}} = \frac{1}{2^{n}}$

$n = number of half-lives$

$∴$ $2^{n} = \frac{A_{0}}{A} = \frac{64}{1} = 2^{6} \Rightarrow n = 6$

$∴$ $Time = n (t_{1 / 2}) = 6 \times t_{1 / 2} = 6 \times 18 = 108 days$

$Hence, after 108 days the laboratory can be considered safe for use.$

Q 2 :

The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by $E = \frac{3}{5} \frac{Z (Z - 1) e^{2}}{4 π ε_{0} R}$

The measured masses of the neutron, ${}_{1}^{1}H$ , ${}_{7}^{15}N$ and ${}_{8}^{15}O$ are $1.008665 u$ , $1.007825 u$ , $15.000109 u$ and $15.003065 u$ , respectively. Given that the radii of both the ${}_{7}^{15}N$ and ${}_{8}^{15}O$ nuclei are same, $1 u = 931.5 MeV / c^{2}$ ( $c$ is the speed of light) and $e^{2} / (4 π ε_{0}) = 1.44 MeV fm$ . Assuming that the difference between the binding energies of ${}_{7}^{15}N$ and ${}_{8}^{15}O$ is purely due to the electrostatic energy, the radius of either of the nuclei is ( $1 fm = 10^{- 15} m$ ). [2016]

2.85 fm
3.03 fm
3.42 fm
3.80 fim

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(3)

Binding energy of nitrogen atom

$[8 M_{n} + 7 M_{p} - M_{N}] \times 931$

$= [8 \times 1.008665 + 7 \times 1.007825 - 15.000109] \times 931$

Binding energy of oxygen atom

$[8 M_{n} + 8 M_{p} - M_{O}] \times 931$

$= [8 \times 1.008665 + 8 \times 1.007825 - 15.003065] \times 931$

$∴$ $Difference = 0.0037960 \times 931 MeV$ $\dots (i)$

$Also$ $E_{O} = \frac{3}{5} \times \frac{8 \times 7}{R} \times \frac{e^{2}}{4 π ε_{0}} = \frac{3}{5} \times \frac{56}{R} \times 1.44 MeV$

$E_{N} = \frac{3}{5} \times \frac{7 \times 6}{R} \times \frac{e^{2}}{4 π ε_{0}} = \frac{3}{5} \times \frac{42}{R} \times 1.44 MeV$

$∴$ $E_{O} - E_{N} = \frac{3}{5} \times \frac{14}{R} \times 1.44 MeV$ $\dots (i i)$

$From eq. (i) & (ii)$

$\frac{3}{5} \times \frac{14}{R} \times 1.44 = 0.0037960 \times 931$

$∴$ $R = 3.42 fm$

Q 3 :

The isotope ${}_{5}^{12}B$ having a mass $12.014 u$ undergoes $β$ -decay to ${}_{6}^{12}C$ . ${}_{6}^{12}C$ has an excited state of the nucleus $({{}_{6}^{12}C}^{})$ at $4.041 MeV$ above its ground state. If ${}_{5}^{12}B$ decays to ${{}_{6}^{12}C}^{}$ , the maximum kinetic energy of the $β$ -particle in units of MeV is ( $1 u = 931.5 MeV / c^{2}$ , where $c$ is the speed of light in vacuum). [2016]

(9)

Here, ${}_{5}^{12}β ⟶ {}_{6}^{12}C + {}_{- 1}^{0}e + \bar{ν}$

Maximum kinetic energy of $β$ -particle

$= [mass of {}_{5}^{12}β - mass of {}_{6}^{12}C] \times 931.5 - 4.041$

$= [12.014 - 12] \times 931.5 - 4.041 = 9 MeV$

Q 4 :

A nuclear power plant supplying electrical power to a village uses a radioactive material of half-life $T$ years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of $n T$ years, then the value of $n$ is [2015]

(3)

$\begin{matrix} N_{0} & \overset{T}{\to} & \frac{N_{0}}{2} & \overset{T}{\to} & \frac{N_{0}}{4} & \overset{T}{\to} & \frac{N_{0}}{8} \\ 100 % & 50 % & 25 % & 12.5 % \end{matrix}$

$So, three half-lives are required.$

$∴$ $t = n (T) = 3 T$

$∴$ $n = 3$

Q 5 :

A fission reaction is given by ${}_{92}^{236}U \to {}_{54}^{140}{Xe} + {}_{38}^{94}{Sr} + x + y,$ where $x$ and $y$ are two particles. Considering ${}_{92}^{236}U$ to be at rest, the kinetic energies of the products are denoted by $K_{Xe}, K_{Sr}, K_{x} (2 MeV)$ and $K_{y} (2 MeV)$ , respectively. Let the binding energies per nucleon of ${}_{92}^{236}U$ , ${}_{54}^{140}{Xe}$ and ${}_{38}^{94}{Sr}$ be $7.5 MeV$ , $8.5 MeV$ and $8.5 MeV$ , respectively. Considering different conservation laws, the correct option(s) is(are) [2015]

$x = n, y = n, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$
$x = p, y = e^{-}, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$
$x = p, y = n, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$
$x = n, y = n, K_{Sr} = 86 MeV, K_{Xe} = 129 MeV$

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(1)

For the given fission reaction,

${}_{92}^{236}U ⟶ {}_{54}^{140}{Xe} + {}_{38}^{94}{Sr} + x + y$

From conservation laws of mass number (A) and atomic number (Z),

$x = n,$ $y = n$ i.e., $x = {}_{0}^{1}n$ and $y = {}_{0}^{1}n$

From conservation of momentum, $| P_{Xe} |$ $=$ $| P_{Sr} |$

From $K = \frac{P^{2}}{2 m} \Rightarrow K \propto \frac{1}{M}$

$∴$ $\frac{K_{Sr}}{K_{Xe}} = \frac{m_{Xe}}{m_{Sr}}$

$∴$ $K_{Sr} = 129 MeV$ and $K_{Xe} = 86 MeV$

Q 6 :

List-I shows various functional dependencies of energy (E) on the atomic number (Z). Energies associated with certain phenomena are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in List-II. [2025]

List-I List-II

(P) $E \propto Z^{2}$ (1) Energy of characteristic X-rays

(Q) $E \propto {(Z - 1)}^{2}$ (2) Electrostatic part of the nuclear binding energy for stable nuclei with mass numbers in the range 30 to 170

(R) $E \propto (Z - 1)$ (3) Energy of continuous X-rays

(S) E is practically independent of Z (4) Average nuclear binding energy per nucleon for stable nuclei with mass number in the range 30 to 170

(5) Energy of radiation due to electronic transitions from hydrogen-like atoms

(P) → (4), (Q) → (3), (R) → (1), (S) → (2)
(P) → (5), (Q) → (2), (R) → (1), (S) → (4)
(P) → (5), (Q) → (1), (R) → (2), (S) → (4)
(P) → (3), (Q) → (2), (R) → (1), (S) → (5)

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(3)

$(P) Energy of H-like atom$ $E = - 13.6 \frac{Z^{2}}{n^{2}}$

$∴$ $E \propto Z^{2}$ $(P) \to (5)$

$(Q) Energy of characteristic X-ray by Moseley's correction$

$E = - 13.6 {(Z - 1)}^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

$or, E \propto {(Z - 1)}^{2}$ $(Q) \to (1)$

$(R) Electrostatic binding energy \propto Z (Z - 1)$

$(R) \to (2)$

$(S) For stable nuclei with mass number in the range 30 to 170,$ $binding energy per nucleon is constant and the graph is a straight line.$

$∴$ $(S) \to (4)$

Q 7 :

Column-II gives certain systems undergoing a process. Column-I suggests changes in some of the parameters related to the system. Match the statements in Column-I to the appropriate process(es) from Column-II. [2009]

Column-I Column-II

(A) The energy of the system is increased (p) System: A capacitor, initially uncharged
Process: It is connected to a battery

(B) Mechanical energy is provided to the system, which is converted into energy of random motion of its parts (q) System: A gas in an adiabatic container fitted with an adiabatic piston
Process: The gas is compressed by pushing the piston

(C) Internal energy of the system is converted into its mechanical energy (r) System: A gas in a rigid container
Process: The gas gets cooled due to colder atmosphere surrounding it

(D) Mass of the system is decreased (s) System: A heavy nucleus, initially at rest
Process: The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted

(t) System: A resistive wire loop
Process: The loop is placed in a time varying magnetic field perpendicular to its plane

A → p, q, t; B → q; C → s; D → s
A → s; B → q; C → s; D → p, q, t
A → s; B → s; C → q; D → p, q, t
A → s; B → p, q, t; C → q; D → s

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(1)

A → p, q, t; B → q; C → s; D → s

(p) When an uncharged capacitor is connected to a battery, it becomes charged and energy is stored $E = \frac{1}{2} Q V$ in the capacitor.

(q) When a gas in an adiabatic container fitted with an adiabatic piston is compressed by pushing the piston:

(i) The internal energy of the system increases

$Δ U = Q - W = 0 - (- P d V) = + P d V$

(ii) Mechanical energy is transferred to the piston, which is converted into the kinetic energy of the gas molecules.

(r) When the gas in a rigid container gets cooled, the internal energy of the system decreases. Due to this, it is converted into mechanical energy.

(s) When a heavy nucleus initially at rest splits into two nuclei of nearly equal masses and some neutrons are emitted, as in case of nuclear fission, internal energy of the system is converted into mechanical energy and some matter is converted into energy.

(t) When a resistive wire loop is placed in a time-varying magnetic field perpendicular to its plane, the energy of the system increases due to the induced current.

Q 8 :

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 a continuous spectrum. Considering a three-body decay process, i.e., $n \to p + e^{-} + {\bar{ν}}_{e},$ around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $({\bar{ν}}_{e})$ 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 $0.8 \times 10^{6} eV$ . The kinetic energy carried by the proton is only the recoil energy.

Q. If the anti-neutrino had a mass of $3 eV / c^{2}$ (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron? [2012]

$0 \leq K \leq 0.8 \times 10^{6} eV$
$3.0 eV \leq K \leq 0.8 \times 10^{6} eV$
$3.0 eV \leq K < 0.8 \times 10^{6} eV$
$0 \leq K < 0.8 \times 10^{6} eV$

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(4)

Kinetic energy $(K)$ of electron will be minimum or zero when total energy is shared by proton and anti-neutrino.

$∴$ $0 \leq K < 0.8 \times 10^{6} eV$

Q 9 :

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 a continuous spectrum. Considering a three-body decay process, i.e., $n \to p + e^{-} + {\bar{ν}}_{e},$ around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $({\bar{ν}}_{e})$ 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 $0.8 \times 10^{6} eV$ . The kinetic energy carried by the proton is only the recoil energy.

Q. What is the maximum energy of the anti-neutrino? [2012]

Zero
Much less than $0.8 \times 10^{6} eV$
Nearly $0.8 \times 10^{6} eV$
Much larger than $0.8 \times 10^{6} eV$

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(3)

$∵$ $K_{p}^{-} + K_{e}^{-} + K_{\bar{ν}}^{-} = 0.8 \times 10^{6} eV$

$When K_{e}^{-} = 0, then K_{p}^{-} + K_{\bar{ν}}^{-} = 0.8 \times 10^{6} eV$

$Mass of \bar{ν} ≪ P$

$∴$ $Maximum energy of anti-neutrino is nearly 0.8 \times 10^{6} eV$

Q 10 :

Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, ${}_{1}^{2}H$ , known as deuteron and denoted by $D$ , can be thought of as a candidate for fusion reactor. The D-D reaction is ${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{3}{He} + n + energy .$ In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ${}_{1}^{2}H$ 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 $t_{0}$ before the particles fly away from the core. If $n$ is the density (number/volume) of deuterons, the product $n t_{0}$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $5 \times 10^{14} s / {cm}^{3}$ . It may be helpful to use the following: Boltzmann constant $k = 8.6 \times 10^{- 5} eV / K; \frac{e^{2}}{4 π ε_{0}} = 1.44 \times 10^{- 9} eVm$ [2009]

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

strong nuclear force acting between the deuterons
coulomb force acting between the deuterons
coulomb force acting between deuteron-electron pairs
the high temperature maintained inside the reactor core

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(4)

In the core of a nuclear fusion reactor, the gas becomes plasma, a collection of ${}_{1}^{2}H$ nuclei and electrons, which is formed due to the high temperature maintained inside the reactor core. High temperature is required for nuclear fusion.

Q 11 :

Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, ${}_{1}^{2}H$ , known as deuteron and denoted by $D$ , can be thought of as a candidate for fusion reactor. The D-D reaction is ${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{3}{He} + n + energy .$ In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ${}_{1}^{2}H$ 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 $t_{0}$ before the particles fly away from the core. If $n$ is the density (number/volume) of deuterons, the product $n t_{0}$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $5 \times 10^{14} s / {cm}^{3}$ . It may be helpful to use the following: Boltzmann constant $k = 8.6 \times 10^{- 5} eV / K; \frac{e^{2}}{4 π ε_{0}} = 1.44 \times 10^{- 9} eV m$ [2009]

Q. Assume that two deuteron nuclei in the core of fusion reactor at temperature $T$ are moving towards each other, each with kinetic energy $1.5 k T$ , 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 $4 \times 10^{- 15} m$ is in the range

$1.0 \times 10^{9} K < T < 2.0 \times 10^{9} K$
$2.0 \times 10^{9} K < T < 3.0 \times 10^{9} K$
$3.0 \times 10^{9} K < T < 4.0 \times 10^{9} K$
$4.0 \times 10^{9} K < T < 5.0 \times 10^{9} K$

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(1)

From conservation of mechanical energy

Loss of kinetic energy of two deuteron nuclei = gain in their potential energy.

$2 \times 1.5 k T = \frac{1}{4 π ε_{0}} \frac{e \times e}{r} = \frac{e^{2}}{4 π ε_{0}} \times \frac{1}{r}$

$\Rightarrow 2 \times 1.5 \times (8.6 \times 10^{- 5} \frac{eV}{K}) \times T = \frac{1.44 \times 10^{- 9} eV m}{4 \times 10^{- 15} m}$

$∴$ $T = \frac{1.44 \times 10^{- 9}}{2 \times 1.5 \times 8.6 \times 10^{- 5} \times 4 \times 10^{- 15}} = 1.4 \times 10^{9} K$

Q 12 :

Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, ${}_{1}^{2}H$ , known as deuteron and denoted by $D$ , can be thought of as a candidate for fusion reactor. The D-D reaction is ${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{3}{He} + n + energy .$ In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ${}_{1}^{2}H$ 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 $t_{0}$ before the particles fly away from the core. If $n$ is the density (number/volume) of deuterons, the product $n t_{0}$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $5 \times 10^{14} s / {cm}^{3}$ . It may be helpful to use the following: Boltzmann constant $k = 8.6 \times 10^{- 5} eV / K; \frac{e^{2}}{4 π ε_{0}} = 1.44 \times 10^{- 9} eV m$ [2009]

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?

deuteron density $= 2.0 \times 10^{12} {cm}^{- 3}$ ,
confinement time $= 5.0 \times 10^{- 3} s$
deuteron density $= 8.0 \times 10^{14} {cm}^{- 3}$ ,
confinement time $= 9.0 \times 10^{- 1} s$
deuteron density $= 4.0 \times 10^{23} {cm}^{- 3}$ ,
confinement time $= 1.0 \times 10^{- 11} s$
deuteron density $= 1.0 \times 10^{24} {cm}^{- 3}$ ,
confinement time $= 4.0 \times 10^{- 12} s$

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(2)

As given in the passage, the product of the deuteron density $(n)$ and confinement time $(t_{0})$ $n t_{0} > 5 \times 10^{14},$ which is the Lawson criterion for a reactor to work successfully.

$∴$ Here $n = 8.0 \times 10^{14} {cm}^{- 3}$ and $t_{0} = 9.0 \times 10^{- 1} s$

Topic Question Set

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	List-I		List-II
(P)	$E \propto Z^{2}$	(1)	Energy of characteristic X-rays
(Q)	$E \propto {(Z - 1)}^{2}$	(2)	Electrostatic part of the nuclear binding energy for stable nuclei with mass numbers in the range 30 to 170
(R)	$E \propto (Z - 1)$	(3)	Energy of continuous X-rays
(S)	E is practically independent of Z	(4)	Average nuclear binding energy per nucleon for stable nuclei with mass number in the range 30 to 170
		(5)	Energy of radiation due to electronic transitions from hydrogen-like atoms

	Column-I		Column-II
(A)	The energy of the system is increased	(p)	System: A capacitor, initially uncharged Process: It is connected to a battery
(B)	Mechanical energy is provided to the system, which is converted into energy of random motion of its parts	(q)	System: A gas in an adiabatic container fitted with an adiabatic piston Process: The gas is compressed by pushing the piston
(C)	Internal energy of the system is converted into its mechanical energy	(r)	System: A gas in a rigid container Process: The gas gets cooled due to colder atmosphere surrounding it
(D)	Mass of the system is decreased	(s)	System: A heavy nucleus, initially at rest Process: The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted
		(t)	System: A resistive wire loop Process: The loop is placed in a time varying magnetic field perpendicular to its plane