Nuclei Test

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Nuclei Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The following nuclear reaction is an example of $$ \frac { 12 }{ 6 } C+\frac { 4 }{ 2 } H\rightarrow \frac { 16 }{ 8 } O+energy$$:

A
Fission

B
Fusion

C
alpha decay

D
beta decay

Solution

It is an example of nuclear fusion reaction. In fusion reaction, two smaller nuclei combine to form a heavy nuclei with release of some energy.
Question 2
1 / -0

$$X(n, \alpha)+_0^1n \rightarrow _{3}^{7}Li + _2^4He$$, then $$X$$ will be :

A
$$_{5}^{10}B$$

B
$$_{5}^{9}B$$

C
$$_{4}^{11}Be$$

D
$$_{2}^{4}He$$

Solution

$$\underbrace { x\left( n,d \right) }_{ _{ a }^{ b }{ x } } +\dfrac { 1 }{ 0 } n\rightarrow \dfrac { 7 }{ 3 } li+\dfrac { 4 }{ 2 } He$$......(given reaction)

$$\displaystyle \sum $$ Atomic number in $$L.H.S=a+0=a$$

$$\displaystyle \sum $$ Atomic number in $$R.H.S=3+2=5$$

$$\therefore \ a=5$$

$$\displaystyle \sum $$ Atomic mass number in $$R.H.S=7-4=11$$

$$\displaystyle \sum $$ Atomic mass number in $$L.H.S=b+1$$

$$\therefore \ b+1=11$$

$$\therefore \ 6=10$$

$$\therefore \ x\to _{ a }^{ b }{ x } \to _{ 5 }^{ 10 }{ x }$$

comparing this with options we get $$_{ 5 }^{ 10 }{ B }$$

Hence option $$A$$ is correct.
Question 3
1 / -0

If $$m_{0}$$ is the mass of isotope $$\frac{16}{8}O, m_{p}\ and\ m_{m}$$ are the masses of a proton and neutron respectively, the nuclear binding energy of the isotope is:

A
$$(m_{0}\ -\ 17m_{n})c^{2}$$

B
$$(m_{0}\ -\ 8m_{p})c^{2}$$

C
$$(m_{0}\ -\ 8m_{p}-8m_{n})c^{2}$$

D
$$m_{0}c^{2}$$

Solution

$$\Delta m = \text{Mass defect=mass of constituent nucleons - mass of molecule}=8m_p +(16-8)m_n -m_0$$
Binding energy =$$\Delta m \times c^2=(8(m_p+m_n)-m_0 )c^2$$
Option C is correct.
Question 4
1 / -0

Calculate the neutron separation energy from the following data.
$$m(^{40}_{20}Ca) = 39.962591 \ u;$$
$$m (^{41}_{20}Ca) = 40.962278 \ u;$$
$$m_n = 1.00865;$$
$$ 1u = 931.5 MeV/C^2$$

A
$$7.57 \space MeV$$

B
$$8.36 \space MeV$$

C
$$9.12 \space MeV$$

D
$$9.56 \space MeV$$

Solution

Energy equation:
$$_{ 20 }^{ 41 }{ Ca }+Q-1\rightarrow_{ 20 }^{ 40 }{ Ca }+_{ 0 }^{ 1 }{ n}$$
Mass defect,
$$\Delta m=m(_{ 20 }^{ 40 }{ Ca })+m(_{ 0 }^{ 1 }{ n })-m(_{ 20 }^{ 41 }{ Ca }\\ \quad=39.962591+1.008665-40.962273\\ \quad=0.008978u$$
BUt
$$1u=931.5MeV.C^2$$
Hence
$$0.008978 u=0.008978\times931.5=8.363MeV$$
Thus separation energy $$=8.363MeV$$
Question 5
1 / -0

Which of the following graph shows variation of decay rate of a radioactive sample with number of nuclei left in the sample :-

A

B

C

D

Solution

Decay rate $$r=-\dfrac{dN}{dt}$$ and $$r=N\lambda$$
Where $$\lambda(decay-constant)$$ is a constant so $$r=\dfrac{-dN}{dt} \propto N$$
Which is a linear relation and $$negative$$ sign denote that as N decrease the rate $$-\dfrac{dN}{dt}$$increase but $$\dfrac{dN}{dt} decreases$$
So option D is correct.(Line with negative slope)
Question 6
1 / -0

Binding energy per nucleon versus mass number curve for nuclei is shown in the figure. A, B, C and D are four muclei indicated on the curve. The process that would release energy is

A
$$ C \rightarrow 2D $$

B
$$A \rightarrow C+D $$

C
$$A \rightarrow 2C $$

D
$$B \rightarrow C +D $$

Solution

We can clearly deduce from the graph that:
(A) $$C > 2D$$
(B) $$C > A\Rightarrow C+D > A$$
(C) $$C > A\Rightarrow 2C > A$$
(D) $$C > B\Rightarrow C+D > B$$
Only in option A, the reactant side has more Binding energy than productant side thus releasing energy in the process.
Option (A) is correct.
Question 7
1 / -0

What is the binding energy of an electron in the first orbit of $${ Li }^{ 2+ }$$ is

A
$$13.6 eV$$

B
$$122.4 eV$$

C
$$-13.6 eV$$

D
$$-122.4 ev$$

Solution

$$ \begin{array}{l} \text { We know, } \\ \text { Energy of electron in Bohr's theory is } \\ \end{array} $$
$$ E=\left(\frac{-13 \cdot 6 z^{2}}{n^{2}}\right)ev $$
$$\begin{aligned} \text { for } & L{i}^{2+} \\ & z=3 \\ & n=1 \end{aligned} $$
$$ \begin{array}{l} E=\frac{-13.6 \times 9}{1} \mathrm{ev} \\ E=-122.4 \mathrm{ev} \end{array} $$
Question 8
1 / -0

The binding energies of two nuclei $$P^{n}$$ and $$Q^{2n}$$ are $$x$$ and $$y$$ joules. If $$2x > y$$ then the energy released in the reaction:
$$P^{n}+P^{n} \rightarrow Q^{2n}$$ will be

A
$$2x+y$$

B
$$2x-y$$

C
$$-(2x-y) $$

D
$$x+y$$

Solution

Given:
Binding energy of nuclei $$P^n=x$$
Binding energy of nuclei $$Q^{2n}=y$$
To find:
Energy released in the reaction.
$$P^n+P^n\rightarrow Q^{2n}$$
Total Binding energy of reactant = $$x+x$$
= $$2x$$
Total Binding energy of product = $$y$$
Energy released in reaction = $$BE_{product}-BE_{reactants}$$
= $$y-2x$$
= $$-(2x-y)$$
Thus, option (C) is correct.
Question 9
1 / -0

If $$N_t=N_0e^{-\lambda t}$$ then number of disintegrated atoms between $$t_1$$ to $$t_2$$ ($$t_2> t_1$$) will be:

A
$$N_0[e^{\lambda t_2}-e^{\lambda t_1}]$$

B
$$N_0[-e^{\lambda t_2}-e^{-\lambda t_1}]$$

C
$$N_0[e^{-\lambda t_1}-e^{-\lambda t_2}]$$

D
none

Solution

Initial amount is $$N_0$$ so amount after time $$t_1$$ will be $$N=N_0e^{-\lambda t_1}$$
And that after time $$t_2$$ is $$N_2=N_0e^{-\lambda t_2 }$$
So decayed in period between $$t_1 $$ and $$t_2$$ is nothing but $$N_1-N_2=N_0(e^{-\lambda t_1}-e^{-\lambda t_2})$$ as $$N_1 \gt N_2$$
option c is correct.
Question 10
1 / -0

The average binding energy per nucleon of a nucleus is of the order of

A
$$8 \,eV$$

B
$$8 \,J$$

C
$$8 \,keV$$

D
$$8 \,MeV$$

Solution