Nuclei Test

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Nuclei Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

What is atomic mass of $$He$$ in amu $$(u)$$ units?

A
$$8$$

B
$$9$$

C
$$2$$

D
$$4$$

Solution

The atomic mass of an element is the average mass of an element measured in an atomic mass unit (amu). The atomic mass of $$He$$ is $$4\ u$$.
Option $$D$$ is correct.
Question 2
1 / -0

Answer the question:
A nuclide has the symbol $$^{22}_{10}Ne,$$
What is the proton number of a nucleus of this nuclide ?

A
$$10$$

B
$$12$$

C
$$22$$

D
$$32$$

Solution

The number of protons in the nucleus of the atom is equal to the atomic number (Z).
Here atomic number is 10.
so proton number of a nucleus of this nuclide is 10.
Hence option A is correct.
Question 3
1 / -0

A stationary body explodes into two fragments each of rest mass 1kg that move apart at speed of 0.6c relative to the original body. The rest mass of the original body is:-

A
2 kg

B
2.5 kg

C
1.6 kg

D
2.25 kg

Solution

Relativistic mass of a body moving with velocity $$v$$ can be given by:
$$m=\dfrac{m_0c^2}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}}$$
Where $$m_0$$= rest mass of particle
Applying energy conservation,
Initial mass energy = Final mass energy
$$m_0c^2=\dfrac{m_{01}c^2}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}}+\dfrac{m_{02}c^2}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}}\Rightarrow m_0=\dfrac{1}{0.8}+\dfrac{1}{0.8}=2.5kg$$
Where $$m_{01}$$= rest mass of particle one fragment = 1 $$kg$$
$$ m_{02} $$ = rest mass of particle two fragment = 1 $$kg$$
$$v$$= velocity of each fragment
Question 4
1 / -0

A radioactive nucleus undergoes a series of decays according to the sequence
$$A\overset{\beta}{\rightarrow}A_1\overset{\alpha}{\rightarrow}A_2\overset{\alpha}{\rightarrow}A_3$$
If the mass number and atomic number of $$A_3$$ are $$172$$ and $$69 $$respectively, then the mass number and atomic number of A is

A
$$56, 23$$

B
$$180, 72$$

C
$$120, 52$$

D
$$84, 38$$

Solution

$$^A_Z{A} \rightarrow A_1 \rightarrow A_2 \rightarrow _69 {A}_3^{172}$$

Mass balance, $$A = 0 + 4 + 4 + 172$$
$$= 180$$
Charge balance, $$Z= -1 + 2 + 2 + 69$$
$$ = 72$$
Question 5
1 / -0

Atomic masses of $$^{14}_{7}N$$ and $$^{16}_{8}O$$ are $$14.008$$ amu and $$16.000$$ amu respectively. The mass of $$^{1}_{1}H$$ atom is 1.007825 amu and the mass of neutron is $$1.008665$$ amu. Pick the correct option

A
$$^{14}_{7}N$$ is more stable than $$^{16}_{8}O$$

B
$$^{16}_{8}O$$ is more stable than $$^{14}_{7}N$$

C
Both are not stable

D
Both are equally stable

Solution

Mass of 7 protons and 7 neutrons=$$7(1.007825+1.008665)=14.11543amu$$
Mass of $$_7^{14}N=14.008amu$$
Hence mass lost as energy in formation of nitrogen atom=$$0.10743amu$$
Mass of 8 protons and 8 neutrons=$$8(1.007825+1.008665)amu=16.13192amu$$
Mass of $$_8^{16}O=16.000amu$$
Hence mass lost as energy in formation of oxygen atom=$$0.13192amu$$
Since more mass has been lost in formation of oxygen atom, oxygen atom is more stable.
Question 6
1 / -0

The mass of chlorine ($$_{17}Cl^{35}$$) atom is 34.98 amu, mass of proton = 1.007825 amu, mass of neutron= 1.008665 amu. Then binding energy is :

A
$$287.83$$ Mev

B
$$287.83$$ joules

C
$$8.22 $$Mev

D
$$8.22 $$joule

Solution

$$ \Delta m = (17 \times 1.007825) + (18 \times 1.008665) - 34.98$$
$$ = 0.308995 amu$$
$$ = 0.308995 \times 931.478 \dfrac{MeV}{c^{2}} ....... [$$using $$1 amu = 931.478 \dfrac{MeV}{c^{2}}]$$
$$ = 287.82 \dfrac{MeV}{c^{2}}$$
$$ E = \Delta m c^{2}$$
$$ = 287.82 MeV$$
Question 7
1 / -0

The energy required to separate the typical middle mass nucleus $$^{120}_{50}Sn$$ into its constituent nucleons is :
(Mass of $$^{120}_{50}Sn=$$ $$119.902199\ amu$$; mass
of proton$$=1.007825\ amu$$ and mass of neutron$$=1.008665\ amu$$)

A
$$921\ MeV$$

B
$$821\ MeV$$

C
$$1021 \ MeV$$

D
$$1121\ MeV$$

Solution

$$Z = 50, A-Z = 120-50 = 70$$

$$\Delta m = Z.m_p + (A-Z)m_n - M$$
$$\Delta m = [50\times 1.007825 + 70\times 1.008665 - 119.902199]$$
$$= 1.095601\ MeV$$

$$E = 1.095601\times 931.478\ MeV$$
$$=1020.53\ MeV$$ $$\approx$$ $$1021\ MeV$$
Question 8
1 / -0

In the reaction $$_7N^{14} +$$ $$ _2He^4 \rightarrow _8O^{17} + $$ $$_1H^1$$. The minimum energy of $$\alpha$$-particle is
$$M_N=$$ $$14.00307$$ amu $$M_{He}=$$ $$4.00260$$ amu
$$M_o =$$ $$16.99914$$ amu $$M_H=$$ $$1.00783$$ amu

A
$$1.21MeV$$

B
$$1.62MeV$$

C
$$1.89MeV$$

D
$$1.96MeV$$

Solution

Mass on reactant side $$= (14.00307 + 4.00260) amu $$

Mass on product side $$ = (16.99914 + 1.00783) amu $$

So, mass defect $$ = 1.3 \times10^{-3} amu.$$

Energy of the $$\alpha $$ particle $$= 931.5\times1.3\times10^{-3} MeV$$

$$ = 1.21 MeV$$
Question 9
1 / -0

Parto of the uranium decay series is shown
$$_{92}U^{238} \rightarrow _{90}Th^{234} \rightarrow _{91}Pa^{234}\rightarrow$$$$_{92}U^{234} \rightarrow _{90}Th^{230}\rightarrow _{88}Ra^{226}$$
How many pairs of isotopes are there in the above series :

A
1

B
2

C
3

D
0

Solution

Atoms of the element having same atomic number and different mass number are known as isotopes.
Thus $$_{92}U^{238}$$ is an isotope of $$_{92}U^{234}$$ whereas $$_{90}Th^{234}$$ is an isotope of $$_{90}Th^{230}$$.
Hence there are two pairs of isotopes.
Question 10
1 / -0

Match list I and list II.
List I List II
A. $$_{1}^{2}H+^{3}_{1}H\rightarrow ^{4}_{2}He +^{1}_{0}n+17.6 MeV$$ 1. Artificial radioactivity

B. $$^{235}_{92}U+^{1}_{0}n\rightarrow^{143}_{56}Ba+^{90}_{36}Kr+3^{1}_{0}n+200Mev$$ 2. Isodiaphers

C. $$^{23}_{13}Al+^{4}_{2}He\rightarrow ^{30}_{14}Si+^{1}_{1}H$$ 3. Atom Bomb

D. $$^{m}_{Z}A\overset{-\alpha}{\rightarrow} ^{m-4}_{Z-2}B$$ 4. Nuclear Fissio 5. Nuclear Fusion

A
A - 5; B - 3,4; C - 1; D - 2

B
A - 2; B - 3,4; C - 1; D - 5

C
A - 1,5; B - 3; C - 4; D - 2

D
A - 3,4; B - 1; C - 5; D - 4,2
Solution
- Converting a stable atom into a suitable radioactive element is called artificial radioactivity.
- Nuclei and its decay product have the same difference of n and p i.e., (n−p)=m−2Z, hence isodiaphers.
- An atom bomb is weapon with great explosion power that results come from sudden release of energy in fission reactions of heavy nuclei (uranium,plutonium ).
- Process of combining two lighter nuclei to give a heavier nucleus is called nuclear fusion.
- Process of splitting of nuclei of heavier atoms into two nuclei of comparable masses is called nuclear fission.
- Thus, A - 5 ; B - 3,4 ; C - 1 ; D - 2

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Nuclei Test - 59

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Nuclei Test - 59

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Question 1

$$8$$

$$9$$

$$2$$

$$4$$

Solution

Question 2

$$10$$

$$12$$

$$22$$

$$32$$

Solution

Question 3

2 kg

2.5 kg

1.6 kg

2.25 kg

Solution

Question 4

$$56, 23$$

$$180, 72$$

$$120, 52$$

$$84, 38$$

Solution

Question 5

$$^{14}_{7}N$$ is more stable than $$^{16}_{8}O$$

$$^{16}_{8}O$$ is more stable than $$^{14}_{7}N$$

Both are not stable

Both are equally stable

Solution

Question 6

$$287.83$$ Mev

$$287.83$$ joules

$$8.22 $$Mev

$$8.22 $$joule

Solution

Question 7

$$921\ MeV$$

$$821\ MeV$$

$$1021 \ MeV$$

$$1121\ MeV$$

Solution

Question 8

$$1.21MeV$$

$$1.62MeV$$

$$1.89MeV$$

$$1.96MeV$$

Solution

Question 9

1

2

3

0

Solution

Question 10

A - 5; B - 3,4; C - 1; D - 2

A - 2; B - 3,4; C - 1; D - 5

A - 1,5; B - 3; C - 4; D - 2

A - 3,4; B - 1; C - 5; D - 4,2

Solution

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