Nuclei(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

NUCLEI MCQs

Total Questions : 30 | Page 1 of 3 pages

If three $^{4} H e$ nuclei combine to form a $^{12} C$ nucleus, energy is liberated. Why can't helium nuclei combine on their own and minimize the energy?

He gas is rare, so collisions between nuclei are also rare;
It happens and the $^{12} C$ we see in nature was formed by this method only in Earth's atmosphere;
The strong nuclear force between nucleons is attractive but short range; the Coulomb force between protons is long range, but repulsive;
Helium nucleus is more stable than carbon.

Discuss Question

Answer: Option C. -> The strong nuclear force between nucleons is attractive but short range; the Coulomb force between protons is long range, but repulsive;
:
C
Nuclear interactions are interesting battle grounds of different forces - mainly the Coulomb force, which is electrostatic in nature, and the strong nuclear force.
The strong nuclear force is a strong attractive force, but only activates when you bring the nucleons closer to each other than a few femtometers. At those length scales, no force in the universe is stronger than the strong nuclear force.
The problem is,

^{4} H e

nuclei are positively charged, because of the two protons. To even bring them to a close distance, we will need to overcome the Coulomb repulsion between them, which increases as separation decreases. Only when they have reached a separation of few fermi the strong nuclear attraction becomes large enough to hold the nucleons together against the Coulomb repulsion.

^{4} H e

nuclei don't automatically fuse to make

^{12} C

because of the energy that needs to be supplied to make them overcome the Coulomb repulsion, before the strong nuclear force even comes to the scene.

Question 2.

1 g of hydrogen is converted into 0.993 g of helium in a thermonuclear reaction. The energy released

$63 \times 10^{7} J$
$63 \times 10^{10} J$
$63 \times 10^{14} J$
$63 \times 10^{20} J$

Discuss Question

Answer: Option B. ->

63 \times 10^{10} J

:
B

Δ m = 1 - 0.993 = 0.007 g m

∴

E = Δ m c^{2} = 0.007 \times 10^{- 3} \times (3 \times 10^{8})^{2} = 63 \times 10^{10} J

Question 3.

The binding energy per nucleon of deuteron $(_{1}^{2} H)$ and helium nucleus $(_{2}^{4} H e)$ is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

13.9 MeV
26.9 MeV
23.6 MeV
19.2 MeV

Discuss Question

Answer: Option C. -> 23.6 MeV
:
C

_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + Q

Total binding energy of helium nucleus =

4 \times 7 = 28 M e V

Total binding energy of each deutron =

2 \times 1.1 = 2.2 M e V

Hence energy released =

28 - 2 \times 2.2 = 23.6 M e V

Question 4.

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of the helium nucleus will be [1 amu= 931 MeV]

28.4 MeV
20.8 MeV
27.3 MeV
14.2 MeV

Discuss Question

Answer: Option A. -> 28.4 MeV
:
A
Helium nucleus consist of two neutrons and two protons.
So binding energy

E = Δ m \times 931 M e V

\Rightarrow E = (2 \times m_{p} + 2 m_{n} - M) \times 931 M e V = (2 \times 1.0073 + 2 \times 1.0087 - 4.0015) \times 931 = 28.4 M e V

Question 5.

$M_{n}$ and $M_{p}$ represent mass of neutron and proton respectively. If an element having nuclear mass M has N-neutron and Z-proton, then the correct relation will be

$M < [N M_{n} + Z M_{p}]$
$M > [N M_{n} + Z M_{p}]$
$M = [N M_{n} + Z M_{p}]$
$M = N [M_{n} + M_{p}]$

Discuss Question

Answer: Option A. ->

M < [N M_{n} + Z M_{p}]

:
A
Actual mass of the nucleus is always less than total mass of nucleons so

M < [N M_{n} + Z M_{p}]

Question 6.

A atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of uranium atom $U^{238}$ is 170 MeV. The number of uranium atoms fission per hour will be

$5 \times 10^{15}$
$10 \times 10^{22}$
$40 \times 10^{21}$
$30 \times 10^{25}$

Discuss Question

Answer: Option C. ->

40 \times 10^{21}

:
C
By using

P = \frac{W}{t} = \frac{n \times E}{t}

where n = Number of uranium atom fission and E = Energy released due to each fission so

300 \times 10^{6} = \frac{n \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{3600} \Rightarrow n = 40 \times 10^{21}

Question 7.

The binding energy per nucleon of $O^{16}$ is 7.97 MeV and that of $O^{17}$ is 7.75 MeV. The energy (in MeV) required to remove a neutron from $O^{17}$ is