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**Previous Years JEE Advanced Questions.**

**Paragraph for Question Nos. 1 to 3**

Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, $_{1}^{2} \mathrm{H}$, known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The

D-D reaction is $_{1}^{2} \mathrm{H}+_{1}^{2} \mathrm{H} \rightarrow_{2}^{3} \mathrm{He}+\mathrm{n}+$ energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of $_{1}^{2} \mathrm{H}$ 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 $\mathrm{t}_{0}$ before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product $\mathrm{nt}_{0}$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $5 \times 10^{14} \mathrm{s} / \mathrm{cm}^{3}$. It may be helpful to use the following :

Boltzmann constant $\mathrm{k}=8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K} ; \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}=1.44 \times 10^{-9} \mathrm{eVm}$

(A) strong nuclear force acting between the deuterons

(B) Coulomb force acting between the deuterons

(C) Coulomb force acting between deuteron-electron pairs

(D) the high temperature maintained inside the reactor core

**[JEE-2009]**

**Sol.**(D)

Due to the high temperature developed as a result of collision & fusion causes the core of fusion reactor to plasma.

(A) $1.0 \times 10^{9} \mathrm{K}<\mathrm{T}<2.0 \times 10^{9} \mathrm{K}$

(B) $2.0 \times 10^{9} \mathrm{K}<\mathrm{T}<3.0 \times 10^{9} \mathrm{K}$

(C) $3.0 \times 10^{9} \mathrm{K}<\mathrm{T}<4.0 \times 10^{9} \mathrm{K}$

(D) $4.0 \times 10^{9} \mathrm{K}<\mathrm{T}<5.0 \times 10^{9} \mathrm{K}$

**[JEE-2009]**

**Sol.**(A) $\left(\frac{3}{2} \mathrm{KT}\right) 2=\frac{\mathrm{Kq}_{1} \mathrm{q}_{2}}{\mathrm{r}} 3 \mathrm{KT}=\frac{\mathrm{Kq}_{1} \mathrm{q}_{2}}{\mathrm{r}}$

$\mathrm{T}=\frac{1.44 \times 10^{-9}}{4 \times 10^{-15} \times 3 \times 8.6 \times 10^{-5}}$

(A) deuteron density $=2.0 \times 10^{12} \mathrm{cm}^{-3},$ confinement

time $=5.0 \times 10^{-3} \mathrm{s}$

(B) deuteron density $=8.0 \times 10^{14} \mathrm{cm}^{-3},$ confinement time $=9.0 \times 10^{-1} \mathrm{s}$

(C) deuteron density $=4.0 \times 10^{23} \mathrm{cm}^{-3},$ confinement time $=1.0 \times 10^{-11} \mathrm{s}$

(D) deuteron density $=1.0 \times 10^{24} \mathrm{cm}^{-3},$ confinement time $=4.0 \mathrm{x} 10^{-12} \mathrm{s}$

**[JEE-2009]**

**Sol.**(B)

(A) zero

(B) much less than $0.8 \times 10^{6} \mathrm{eV}$

(C) Nearly $0.8 \times 10^{6} \mathrm{eV}$

(D) Much larger than $0.8 \times 10^{6} \mathrm{eV}$

**[JEE 2012]**

**Sol.**(C)

**Paragraph for Questions 5 and 6**

The mass of a nucleus $_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}$ 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 $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ only if $\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)$ < M. Also two light nuclei of masses $\mathrm{m}_{3}$ and $\mathrm{m}_{4}$ can undergo complete fusion and form a heavy nucleus of mass M’ only if $\left(\mathrm{m}_{3}+\mathrm{m}_{4}\right)$ > M’. The masses of some neutral atoms are given in the table below :-

84 $P O$ at rest undergoes alpha decay, is :-

(A) 5319 (B) 5422 (C) 5707 (D) 5818

**[JEE Advance-2013]**

**Sol.**(A)

(A) The nucleus $_{3}^{6}$ Li can emit an alpha particle

(B) The nucleus $_{84}^{210}$ Po can emit a proton

(C) Deuteron and alpha particle can undergo complete fusion

(D) The nuclei $_{30}^{70} \mathrm{Zn}$ and $_{34}^{82} \mathrm{Se}$ can undergo complete fusion

**[JEE Advance-2013]**

**Sol.**(C)

**[JEE Advance-2013]**

**Sol.**(C)

Considering $_{92}^{236} \mathrm{U}$ to be at rest, the kinetic energies of the products are denoted by

$\mathrm{K}_{\mathrm{Xe}}, \mathrm{K}_{\mathrm{sr}}, \mathrm{K}_{\mathrm{x}}(2 \mathrm{MeV})$ and $\mathrm{K}_{\mathrm{v}}(2 \mathrm{MeV}),$ respectively. Let the binding energies per nucleon of

**[JEE Advance-2015]**

**Sol.**(A)

the nucleus $\left(_{6}^{12} \mathrm{C}^{*}\right)$ at $4.041 \mathrm{MeV}$ above its ground state. If 12

5 $\mathrm{decays}$ to $_{6}^{12} \mathrm{C}^{*},$ the maximum

kinetic energy of the $\beta$ -particle in units of MeV is $\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}\right.$, where c is the speed of light in vacuum).

**[JEE Advance-2016]**

**Sol.**9

(A) 2.85 fm (B) 3.03 fm (C) 3.42 fm (D) 3.80 fm

**[JEE Advance-2016]**

**Sol.**(C)

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