Nuclei Test

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Nuclei Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

$$^{23}Na$$ is the more stable isotope of $$Na$$. By which process $$^{24}_{11}Na$$ can undergo radioactive decay?

A
$$\beta^--emission$$

B
$$\alpha-emission$$

C
$$\beta^+-emission$$

D
None of these

Solution

To convert $$^{24}_{11}Na$$ to $$^{23}_{11} Na$$, we need to decrease the mass number of $$^{24}_{11}Na$$ by $$1$$.
Since, in $$\beta^{-}$$ emission and $$\beta^{+}$$ emission, the mass number of product nucleus is same as that of parent nucleus.
Whereas, in $$\alpha$$ emission, the mass number of product nucleus gets decreased by $$4$$ compared to that of parent nucleus.
Thus $$^{24}_{11} Na$$ cannot undergo any of the radioactive decay to convert into $$^{23}_{11}Na$$.
Question 2
1 / -0

Directions For Questions

Some of the heavy nuclei decay into lighter nuclei by emitting a-particle and acquire stability. The principles of conservation of atomic number, mass number, linear momentum and energy (including rest mass energy) are valid for these decays.
A stationary nucleus $$^{226}_{88}Ra$$ (ground state) decays into the nucleus $$^{222}_{86}Rn$$ (ground state) by emitting an $$\alpha - particle$$.
Given that $$m(^{226}_{88}Ra)=226.02540 u$$. $$m(^4_2He)=4.00260 u$$. $$m(^{222}_{86}Rn)=222.01750 u$$.

...view full instructions

The K.E. of the emitted $$\alpha-particle$$ in the decay of $$^{226}_{88}Ra$$ (approximately)

A
2.3 MeV

B
4.85 MeV

C
9.7 MeV

D
14 MeV

Solution

$$ \begin{array}{c} _{88}^{226} Ra \longrightarrow{ }_{2}^{4} He^{2+}+{ }_{86}^{222} Rn \end{array} $$
$$ 226.02540 \mu \quad 4.00260 \mu \quad 222.01750 \mu $$
$$ \begin{aligned} \Delta m &=0.0053 \\ \text { Energy } &=0.0053 \times 931.5 \mathrm{MeV} \\ &=4.94 \\ K_{1} &=\frac{m_{2}}{m_{1}+m_{2}} \times K \\ &=\frac{4}{222+4} \times 4.94 \\ &=4.8495 \end{aligned} $$
Question 3
1 / -0

In the nuclear reaction
$$^{14}_7N\, +\, _2^4He\, \rightarrow\, ^{17}_8O\, +\, ...............$$
What is the missing particle ?

A
$$_2^4He$$

B
$$_1^1H$$

C
$$_0^1n$$

D
$$\gamma$$

Solution

The hydrogen isotope is missing with atomic number 1 and mass number 1.
Question 4
1 / -0

Fusion reaction is also known as ..........

A
Chemical reaction

B
Elastic reaction

C
Thermonuclear reaction

D
Photonuclear reaction

Solution

In nuclear physics,nuclear fusion is a nuclear reaction in which two or more atomic nuclei collide at very high speed and join to form a new type of atomic nucleus. During this process matter is not conserved because some of the matter of fusing nuclei is converted to photons also known as thermonuclear reaction.
Question 5
1 / -0

Pick out the isoelectronic species from the following :
$$(I)\, {CH_{3}}^{+}\, \quad\, (II)\, H_3O^+\, \quad\, (III)\, NH_3\, \quad\, (IV)\, {CH_{3}}^{-}$$

A
I and II

B
III and IV only

C
I and III

D
II, III, IV

Solution

Isoelectronic species means the no of electron in both molecule is same.
No. of electron in (I) is 9, in (II)=10, (III)=10, (IV)=10.
So II, III, IV are isoelectronic species.
Question 6
1 / -0

When a hydrogen bomb explodes, which of the following is used?

A
Fission

B
Fussion

C
Both

D
None of these

Solution

Hydrogen bomb is based on nuclear fusion and hence hydrogen atoms combine to form Helium atoms.
Question 7
1 / -0

Two atoms of the same element are found to have different number of neutrons in their nuclei. These two atoms are :

A
Isomers

B
lsotopes

C
Isobars

D
Allotropes

Solution

(See all the isotopes of carbon have different no of neutrons.)
Isotopes are atoms with the same number of protons but that have a different number of neutrons. Since the atomic number is equal to the number of protons and the atomic mass is the sum of protons and neutrons, we can also say that isotopes are elements with the same atomic number but different mass numbers.
Question 8
1 / -0

The isotopes present in naturally occurring uranium are

$$(i)$$ $$U-235$$
$$(ii)$$ $$U-236$$
$$(iii)$$ $$U-238$$
$$(iv)$$ $$U-232$$

A
$$(i)$$ and $$(iii)$$

B
$$(i)$$ and $$(ii)$$

C
$$(ii)$$ and $$(iii)$$

D
$$(iv)$$ and $$(ii)$$

Solution

U-235 and U-238 occurs naturally.
Question 9
1 / -0

Which atom contains exactly 15 neutrons?

A
$$P^{32}$$ (atomic number $$= 17$$)

B
$$S^{32}$$ (atomic number $$= 16$$)

C
$$O^{15}$$ (atomic number $$= 8$$)

D
$$N^{15}$$ (atomic number $$= 7$$)

Solution

$$mass\space number= no. of \space protons + no. of\space neutrons$$
Given that mass number of P = 32
Atomic number (no. of electrons $$=$$ no. of protons) $$=$$ 17
Number of neutrons $$= 32 - 17 = 15$$
Hence, P satisfies the requirement.
Question 10
1 / -0

Calculate the binding energy per nucleon for Beryllium $$_{4}{Be}^{9}$$, its mass being $$9.012 \ u$$. The masses of proton and neutron are $$1.008 u$$ and $$1.009 u$$. (Take $$1 u = 931.5 MeV$$)

A
8.3636 MeV

B
7.2828 MeV

C
9.4444 MeV

D
6.7275 MeV

Solution

First, let us find the mass defect. The Beryllium nucleus contains 4 protons and 5 neutrons.
Mass of 4 protons $$= 4 \times 1.008 = 4.032 u$$; Mass of 5 neutrons $$= 5 \times 1.009 u = 5.045 u$$
Total mass of protons (4) and neutrons (5) $$= 4.032 + 5.045 = 9.077 u$$
Mass defect $$= 9.077 - 9.012 = 0.065 u$$
The mass defect converted into equivalent energy gives binding energy.
$$1 u = 931.5 MeV$$
$$\therefore 0.065 u = 0.065 \times 931.5 MeV = 60.5475 MeV$$
Binding energy per nucleon $$= \displaystyle\frac{Binding energy}{Number of nucleons} = \displaystyle\frac{60.5475}{9} = 6.7275 MeV$$

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Nuclei Test - 36

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Nuclei Test - 36

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SHARING IS CARING

Question 1

$$\beta^--emission$$

$$\alpha-emission$$

$$\beta^+-emission$$

None of these

Solution

Question 2

2.3 MeV

4.85 MeV

9.7 MeV

14 MeV

Solution

Question 3

$$_2^4He$$

$$_1^1H$$

$$_0^1n$$

$$\gamma$$

Solution

Question 4

Chemical reaction

Elastic reaction

Thermonuclear reaction

Photonuclear reaction

Solution

Question 5

I and II

III and IV only

I and III

II, III, IV

Solution

Question 6

Fission

Fussion

Both

None of these

Solution

Question 7

Isomers

lsotopes

Isobars

Allotropes

Solution

Question 8

$$(i)$$ and $$(iii)$$

$$(i)$$ and $$(ii)$$

$$(ii)$$ and $$(iii)$$

$$(iv)$$ and $$(ii)$$

Solution

Question 9

$$P^{32}$$ (atomic number $$= 17$$)

$$S^{32}$$ (atomic number $$= 16$$)

$$O^{15}$$ (atomic number $$= 8$$)

$$N^{15}$$ (atomic number $$= 7$$)

Solution

Question 10

8.3636 MeV

7.2828 MeV

9.4444 MeV

6.7275 MeV

Solution

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