Nuclei Test

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Nuclei Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The binding energies of the atoms of elements $$P$$ and $$Q$$ are $$E_{p}$$ and $$E_{Q}$$, respectively. Three atoms of element $$Q$$ fuse to form one atom of element $$P$$. In this process, the energy released is $$e$$. The correct relation between $$E_{P}, E_{Q}$$ and $$e$$ will be

A
$$E_{Q}=3E_{P}+e$$

B
$$E_{Q}=3E_{P} -e$$

C
$$E_{P}=3E_{Q} +e$$

D
$$E_{P}=3E_{Q}-e$$

Solution

Binding energy of $$3$$ $$Q$$ elements will be $$3E_Q$$
So, by energy conservation, binding energy of three $$P$$ elements is equal to the sum of $$Q$$ element's binding energy and the released energy.
So, equation is
$$E_p + e = 3E_Q $$
$$E_p = 3E_Q - e $$
Question 2
1 / -0

In nuclear reaction $$_{4}Be^{9}+_{2}He^{4} \rightarrow _{6}c^{12}$$+X, X will be :

A
Proton

B
Neutron

C
$$\beta $$ -particle

D
$$\alpha $$ -particle

Solution

$$_{4}Be^{9} + _{2}He^{4} \rightarrow _{6}C^{12} + X$$

No. of protons are already balanced with $$C$$.
$$\therefore$$ $$X$$ carries no charge.
and $$ 9+4 > 12$$ by $$1$$ unit mass.
$$\therefore$$ $$X$$ carries one unit mass.
It's a neutron.
Question 3
1 / -0

Among the following reactions , the impossible one is :

A
$$^2He_4+$$ $$^4Be_9\longrightarrow$$ $$^0n_1+$$ $$^6C_{12}$$

B
$$^2He_4 +$$ $$^7N_{14} \longrightarrow$$ $$^1H_1+$$ $$^8O_{17}$$

C
$$4(^1H_1) \longrightarrow$$ $$^2He_4+$$ $$^2(^{-1}e_0)$$

D
$$^3Li_7+$$ $$^1H_1\longrightarrow$$ $$^4Be_8$$

Solution

For possible reactions :
Sum of mass numbers LHS should be equal to the sum of mass numbers on RHS of the equation.
As well as sum of Atomic numbers on LHS should be equal to the sum of Atomic numbers on RHS of the equation.
So, in option C we can see that on LHS mass number is 4 x 1 = 4 and Atomic number is 4 x 1 = 4. on RHS mass number is 2 + 2(-1) = 0 and Atomic number is 4 + 0 = 4.
So, mass number are different in LHS and RHS of the equation. So, this reaction is not possible.
Question 4
1 / -0

On the bombardment of Boron with neutron, an $$\alpha $$ - particle is emitted and product nucleus formed is $$\underline{\hspace{0.5in}}$$

A
$$_{6}C^{12}$$

B
$$_{2}Li^{6}$$

C
$$_{3}Li^{7}$$

D
$$_{4}Be^{9}$$

Solution

$$ _{5}B^{10}+^{1}n \rightarrow _{Z}^{A}X + _{2}^{4}He$$

Mass balance: $$ 10 + 1 = A + 4$$
Atomic no balance: $$ 5 + 0 = Z + 2$$
$$ Z = 3$$
Therefore, it's $$ _{3}Li^{7}$$
Question 5
1 / -0

The particle A is converted to C via following reactions then :
$$A \rightarrow B + _{2}He^{4}$$
$$B \rightarrow C + 2 _{-1}e^{0}$$

A
A and C are isobars

B
A and C are isotopes

C
A and B are isobars

D
A and B are isotopes

Solution

No of neutrons $$ =                  N_A             N_B           N_C$$

No of protons $$ =                      Z_A             Z_B             Z_C$$

Mass $$ =                   N_A+Z_A N_B+Z_B    N_C+Z_C$$

For reaction $$ A \rightarrow   B + _{2}He^{4}$$
                     $$ Z_A = Z_B + 2    ............. [1] $$

For reaction $$B \rightarrow C + 2 _{-1}e^{0}$$
                   $$ Z_B = Z_C - 2 ..............[2]$$


Solving these two equations, we get:
                     $$ Z_A = Z_C$$

Hence they are isotopes.
Question 6
1 / -0

The missing particle in the reaction :
$$^{253}_{99}Es+^{4}_{2}He\rightarrow ^{256}_{101}Md+ \underline{ \hspace{0.5in}}$$

A
deuteron

B
proton

C
neutron

D
$$\beta $$ - particle

Solution

Let the particle be $$X$$.
On summing the mass number on left hand side we get $$253+4 = 257$$ units.
On right hand side we have $$256$$ units only. So, we should have $$1$$ unit mass number on $$X$$ element.
On summing the atomic number on left hand side we get $$99 + 2 = 101$$ units
On right hand side we have $$101$$ units of atomic number. So, we should have $$0$$ unit atomic number on $$X$$ element.
So $$^1_0X = ^1_0n$$ i.e. neutron.
Question 7
1 / -0

Two deuterons combine to form a tritium nucleus and a __

A
$$\alpha $$-particle

B
Fast neutron

C
Proton

D
$$\beta $$ -particle

Solution

The given nuclear reaction is-
$$^2_1H + ^2_1H \rightarrow ^3_1H + ^A_ZX$$
where $$A$$ and $$Z$$ are the mass number and atomic number of element $$X$$, respectively.
Conservation of mass number : $$2+2 = 3+ A$$
$$\implies$$ $$A =1$$
Conservation of atomic number : $$1+1 = 1+ Z$$
$$\implies$$ $$Z =1$$
Thus the element $$X$$ is a proton i.e. $$^1_1H$$.
Question 8
1 / -0

Nuclear fission and fusion can be explained on the basis of Einstein's :

A
Theory of relativity

B
Specific heat equation

C
Mass-energy relation

D
Photoelectric equation

Solution

In nuclear fission and fusion rest mass of reactants are converted into energy. So, it is best explained on the basis of Einstein mass-energy relation.
Einstein theory of relativity deals with relative speed and time.
Einstein specific heat equation is about heat.
Einstein photo-electric equation is on the photo-electric generation.
i.e. generation of electricity due to light.
Question 9
1 / -0

The energy required to remove one neutron from $$_{13}Al^{27}$$ is
(given mass of $$_{13}Al^{27}=$$26.981541 amu, mass of $$_{13}Al^{26}=$$25.984895 amu, mass of neutron $$=$$ 1.008665 amu)

A
6.525MeV

B
11.195MeV

C
4.232MeV

D
8.464MeV

Solution

$$^{27}_{13}Al \rightarrow ^{23}_{13}Al + ^1n$$
$$BE = [25.984895 + 1.008665 - 26.981541][931.478]\ MeV$$
$$= 11.195\ MeV$$
Question 10
1 / -0

The BE/A (binding energy per atomic mass) for deuteron and an $$\alpha $$ particle are X$$_{1}$$and X$$_{2}$$ respectively. The energy released in the reaction will be :

A
X$$_{2}$$-X$$_{1}$$

B
2(X$$_{2}$$-X$$_{1}$$)

C
4(X$$_{2}$$-X$$_{1}$$)

D
8(X$$_{2}$$-X$$_{1}$$)

Solution

$$ \dfrac{BE}{A}]_{_{1}^{2}D} = X_{1} , \dfrac{BE}{A}]_{\alpha} = X_{2}$$
$$ {BE}]_{D} = 2X_{1} $$
$$ {BE}]_\alpha = 4X_{2}$$

$$ 2 _{1}^{2}D \rightarrow _{2}^{4}He$$

Energy released $$= BE]_{product} - BE]_{reactant}$$
$$ = 4X_{2} - 2(2X_{1})$$
$$ = 4(X_{2}-X_{1})$$

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Nuclei Test - 22

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Nuclei Test - 22

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Question 1

$$E_{Q}=3E_{P}+e$$

$$E_{Q}=3E_{P} -e$$

$$E_{P}=3E_{Q} +e$$

$$E_{P}=3E_{Q}-e$$

Solution

Question 2

Proton

Neutron

$$\beta $$ -particle

$$\alpha $$ -particle

Solution

Question 3

$$^2He_4+$$ $$^4Be_9\longrightarrow$$ $$^0n_1+$$ $$^6C_{12}$$

$$^2He_4 +$$ $$^7N_{14} \longrightarrow$$ $$^1H_1+$$ $$^8O_{17}$$

$$4(^1H_1) \longrightarrow$$ $$^2He_4+$$ $$^2(^{-1}e_0)$$

$$^3Li_7+$$ $$^1H_1\longrightarrow$$ $$^4Be_8$$

Solution

Question 4

$$_{6}C^{12}$$

$$_{2}Li^{6}$$

$$_{3}Li^{7}$$

$$_{4}Be^{9}$$

Solution

Question 5

A and C are isobars

A and C are isotopes

A and B are isobars

A and B are isotopes

Solution

Question 6

deuteron

proton

neutron

$$\beta $$ - particle

Solution

Question 7

$$\alpha $$-particle

Fast neutron

Proton

$$\beta $$ -particle

Solution

Question 8

Theory of relativity

Specific heat equation

Mass-energy relation

Photoelectric equation

Solution

Question 9

6.525MeV

11.195MeV

4.232MeV

8.464MeV

Solution

Question 10

X$$_{2}$$-X$$_{1}$$

2(X$$_{2}$$-X$$_{1}$$)

4(X$$_{2}$$-X$$_{1}$$)

8(X$$_{2}$$-X$$_{1}$$)

Solution

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