1
IIT-JEE 2009 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

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 $$E = {{{p^2}} \over {2m}}$$. Thus, the energy of the particle can be denoted by a quantum number 'n' taking values 1, 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 $$h = 6.6 \times {10^{ - 34}}$$ J-s and $$e = 1.6 \times {10^{ - 19}}$$ C.

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

A
$${n^{ - 3/2}}$$
B
$${n^{ - 1}}$$
C
$${n^{1/2}}$$
D
$$n$$
2
IIT-JEE 2009 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

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 $$nt_0$$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 $$\times$$ 10$$^{14}$$ s/cm$$^3$$.

It may be helpful to use the following : Boltzmann constant $$k = 8.6 \times {10^{ - 5}}$$ eV/K; $${{{e^2}} \over {4\pi {\varepsilon _0}}} = 1.44 \times {10^9}$$ eVm.

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

A
strong nuclear force acting between the deuterons.
B
Coulomb force acting between the deuterons.
C
Coulomb force acting between deuteron-electrons pairs.
D
the high temperature maintained inside the reactor core.
3
IIT-JEE 2009 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

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 $$nt_0$$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 $$\times$$ 10$$^{14}$$ s/cm$$^3$$.

It may be helpful to use the following : Boltzmann constant $$k = 8.6 \times {10^{ - 5}}$$ eV/K; $${{{e^2}} \over {4\pi {\varepsilon _0}}} = 1.44 \times {10^9}$$ eVm.

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 $$\times$$ 10$$^{-15}$$ m is in the range

A
$$1.0 \times {10^9}K < T < 2.0 < {10^9}K$$
B
$$2.0 \times {10^9}K < T < 3.0 < {10^9}K$$
C
$$3.0 \times {10^9}K < T < 4.0 < {10^9}K$$
D
$$4.0 \times {10^9}K < T < 5.0 < {10^9}K$$
4
IIT-JEE 2009 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

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 $$nt_0$$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 $$\times$$ 10$$^{14}$$ s/cm$$^3$$.

It may be helpful to use the following : Boltzmann constant $$k = 8.6 \times {10^{ - 5}}$$ eV/K; $${{{e^2}} \over {4\pi {\varepsilon _0}}} = 1.44 \times {10^9}$$ eVm.

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?

A
Deuteron density = $$2.0\times10^{12}~\mathrm{cm^{-3}}$$; Confinement time = $$5.0\times10^{-3}~\mathrm{s}$$.
B
Deuteron density = $$8.0\times10^{14}~\mathrm{cm^{-3}}$$; Confinement time = $$9.0\times10^{-1}~\mathrm{s}$$.
C
Deuteron density = $$4.0\times10^{23}~\mathrm{cm^{-3}}$$; Confinement time = $$1.0\times10^{-11}~\mathrm{s}$$.
D
Deuteron density = $$1.0\times10^{24}~\mathrm{cm^{-3}}$$; Confinement time = $$4.0\times10^{-12}~\mathrm{s}$$.
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