Nuclei Test

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Nuclei Test - 28

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Weekly Quiz Competition

Question 1
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When four hydrogen nuclei fuse together to form a helium nucleus, then in this process

A
energy is absorbed.

B
energy is liberated.

C
absorption and liberation of energy depends upon the temperature.

D
energy is neither liberated nor absorbed.

Solution

In the basic hydrogen fusion cycle four hydrogen nuclei (photons) come together to make a helium nucleus. This fusion cycle releases energy in the core of the star i.e., energy is liberated in this process
Question 2
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What is the mass of one atom of $$C-12$$ in grams?

A
1.992$$\displaystyle \times 10^{-23}$$ gm

B
1.989$$\displaystyle \times 10^{-23}$$ gm

C
1.892$$\displaystyle \times 10^{-23}$$ gm

D
1.965$$\displaystyle \times 10^{-23}$$ gm

Solution

Mass of $$1$$ mole = $$12\,gm$$

Mass of $$6.022 \times 10^{23}\,atom = 12\,gm$$

mass of $$1$$ atom = $$\dfrac{12}{6.023 \times 10^{23}} = 1.993 \times 10^{-23}\,gm$$
Question 3
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Isotopes are the atoms of the same element which contain equal number of

A
nucleons

B
neutrons

C
protons

D
neutrons and protons

Solution

Isotopes of the element are atoms of the element that have the same number of protons, electrons but different number of neutrons.
Option C is correct.
Question 4
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Directions For Questions

An $$\alpha$$-particle passes through a potential difference of $$2 \times 10^{6}$$ volt and then it becomes incident on a silver foil. The atomic number of silver is $$47$$.

...view full instructions

The kinetic energy of $$\alpha$$-particles at a distance $$5 \times 10^{-14}$$m from the nucleus will be(in Joules)

A
$$6.4 \times 10^{-13}$$

B
$$4.3 \times 10^{-13}$$

C
$$2.1 \times 10^{-13}$$

D
$$3.4 \times 10^{-14}$$

Solution

The charge on an alpha particle is $$+2$$.
Hence $$q = 2e = 3.2 \times 10^{-19} C$$
This alpha particle is accelerated by a potential of $$2 \times 10^{6} V$$
Hence Energy of the particle $$ = qV = 3.2 \times 10^{-19} \times 2 \times 10^{6} = 6.4 \times 10^{-13} J$$
This is the total energy.
Potential energy when it is at a distance of $$5 \times 10^{-14} m $$ from the nucleus
$$PE = \dfrac{1}{4\pi \epsilon_{0}}\dfrac{q_{1}q_{2}}{d}$$
Nucleus contains 47 protons.
Hence, $$q_{2} = 47e = 47 \times 1.6 \times 10^{-19} C$$
Substituting these values:
We get PE = $$ 4.3 \times 10^{-13} J$$
$$KE + PE = TE$$
$$KE = TE - PE = (6.4 - 4.3) \times 10^{-13} J = 2.1 \times 10^{-13}J$$
Question 5
1 / -0

The energy released per fission of Uranium is 200 MeV. Determine the number of fission per second required to generate 2MW power.

A
$$6.25\times 10^{16}$$

B
$$0.25\times 10^{16}$$

C
$$1.25\times 10^{16}$$

D
$$25\times 10^{16}$$

Solution

Energy obtained per fission
$$=200 MeV=200\times 1.6\times 10^{-13}J=3.2\times 10^{-11}J$$
no. of fission per second required
$$=\frac {2\times 10^6}{3.2\times 10^{-11}}=6.25\times 10^{16}$$.
Question 6
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Write the following steps of the liquid drop model in sequential order to show the various stages of a nuclear fission reaction.
(a) During the bombardment of a nucleus by neutrons, a neutron could be absorbed by the drop like the nucleus.
(b) A heavy nucleus is treated as the drop of a liquid with the positive charge uniformly distributed.
(c) This leads to the disturbance of the internal forces and sets up oscillations.
(d) From this stage the repulsive forces take over causing the fission of the nucleus.
(e) These oscillations lead to the development of a thick neck forming two parts, giving the nucleus and elongated shape.

A
b c e a d

B
a c e d b

C
a d c e b

D
b a c e d

Solution

The liquid drop model is the theoretical work of Bohr and Wheeler in 1939. The various stages of nuclear fission reaction is nucleus is like +ve charged drop of liquid and when we bombarded neutrons, drop absorbed all of them. So internal forces get disturbed which makes oscillations and drop get divided in two parts. Which results nuclear fission reaction.
Hence option 'D' is correct.
Question 7
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One a.m.u. or one 'u' is equal to :

A
$$1.66053892 10^{-27} kg$$

B
$$1.66053892 10^{-29} kg$$

C
$$1.23457656 10^{-27} kg$$

D
$$1.23457656 10^{-29} kg$$

Solution

One a.m.u.(atomic mass unit) or one 'u' is $$\dfrac{1}{12}$$ of the mass of one carbon-$$12$$ atom.
It is equal $$1.6605389210 \times 10^{-24} g $$ or $$1.6605389210 \times 10^{-27} kg$$.
Question 8
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The sun's energy is due to

A
the nuclear fission of hydrogen.

B
the nuclear fusion of hydrogen.

C
the nuclear fusion of uranium.

D
the nuclear fission of uranium.

Solution

$$^1H_1+^1H_1\rightarrow ^2H_1+e^++\nu$$
So, $$(B)$$ is correct option
Question 9
1 / -0

The modern atomic mass unit is based on the mass of :

A
C-12 isotope

B
hydrogen

C
oxygen

D
nitrogen

Solution

One a.m.u. or one 'u' is equal to $$1.66053892 10^{-24}\ g $$ or $$1.66053892 10^{-27}\ kg$$. It is equal to $$\dfrac{1}{12}$$ of the mass of an atom of carbon-12.

It is used as a standard. The masses of all other atoms are determined relative to the mass of an atom of carbon-12.
Question 10
1 / -0

The average $$\ KE $$ of molecules in a gas at temperature $$\ T $$ is $$\displaystyle \frac {3}{2}kT $$. Find the temperature at which the average KE of molecules equal to binding energy of its atoms.

A
$$\ 1.05 \times 10^{5}K $$

B
$$\ 1.05 \times 10^{4}K $$

C
$$\ 1.05 \times 10^{3}K $$

D
none of these

Solution

Binding energy of atoms $$=13.6\ eV=2.1789\times 10^{-18}J$$
Now since this energy is equal to $$\dfrac{3}{2}kT$$
$$T=\dfrac{2}{3}\dfrac{2.1789\times10^{-18}}{k}=1.05\times 10^5K$$

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Nuclei Test - 28

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Nuclei Test - 28

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Question 1

energy is absorbed.

energy is liberated.

absorption and liberation of energy depends upon the temperature.

energy is neither liberated nor absorbed.

Solution

Question 2

1.992$$\displaystyle \times 10^{-23}$$ gm

1.989$$\displaystyle \times 10^{-23}$$ gm

1.892$$\displaystyle \times 10^{-23}$$ gm

1.965$$\displaystyle \times 10^{-23}$$ gm

Solution

Question 3

nucleons

neutrons

protons

neutrons and protons

Solution

Question 4

$$6.4 \times 10^{-13}$$

$$4.3 \times 10^{-13}$$

$$2.1 \times 10^{-13}$$

$$3.4 \times 10^{-14}$$

Solution

Question 5

$$6.25\times 10^{16}$$

$$0.25\times 10^{16}$$

$$1.25\times 10^{16}$$

$$25\times 10^{16}$$

Solution

Question 6

b c e a d

a c e d b

a d c e b

b a c e d

Solution

Question 7

$$1.66053892 10^{-27} kg$$

$$1.66053892 10^{-29} kg$$

$$1.23457656 10^{-27} kg$$

$$1.23457656 10^{-29} kg$$

Solution

Question 8

the nuclear fission of hydrogen.

the nuclear fusion of hydrogen.

the nuclear fusion of uranium.

the nuclear fission of uranium.

Solution

Question 9

C-12 isotope

hydrogen

oxygen

nitrogen

Solution

Question 10

$$\ 1.05 \times 10^{5}K $$

$$\ 1.05 \times 10^{4}K $$

$$\ 1.05 \times 10^{3}K $$

none of these

Solution

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