Nuclei Test

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Nuclei Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

During a $$\beta$$ positive decay experiment it is observed that kinetic energy of $$\beta$$ positive particle is 50% of Q value of the reaction, then select the correct alternative

A
Almost remaining 50% energy must be in form of kinetic energy of daughter nucleus.

B
Almost remaining 50% energy must be lost in form of heat.

C
Almost remaining 50% energy must be taken away by neutrino.

D
Almost remaining 50% energy must be taken away by antineutrino.

Solution

$$_{Z}X^{A}\rightarrow _{Z-1}Y^{A}+_{+1}e^{0}+v+Q$$
Almost all energy came from Q value is distributed between $$_{+1}e^{0}$$ & $$v.$$
Question 2
1 / -0

When the number of nucleons in nuclei increase, the binding energy per nucleon

A
increases continuously with mass number

B
decreases continuously with mass number

C
remains constant with mass number

D
first increases and then decreases with increase of mass number

Solution

The binding energy per nucleon first increases up to iron-56, , then decreases gradually as shown in the figure.
As Z increases the number of nucleons in nuclei increase. Hence the binding energy per nucleon first increases and then decreases with increase of mass number.
Question 3
1 / -0

A beam of $$16 MeV$$ deutrons from a cyclotron falls on a copper block. The beam is equivalent to a current of 15$$\mu A$$. At what rate do the deutrons strike the block?

A
$$9.4 \times 10^9$$

B
$$9.4 \times 10^7$$

C
$$9.4 \times 10^{11}$$

D
$$9.4 \times 10^{13}$$

Solution

Given: $$I = 15\mu A$$ $$=1.5\times 10^{-5}A$$ and $$|e|=1.6\times 10^{-19}C$$

Using, $$I = \dfrac{q}{t}$$ where $$q=n|e|$$
$$\therefore$$ Rate of deutrons striking the block, $$\dfrac{dn}{dt} = \dfrac{I}{|e|} =\dfrac{1.5\times 10^{-5}}{1.6\times 10^{-19}} = 9.4\times 10^{13}$$
Question 4
1 / -0

In an $$\alpha-decay$$, the kinetic energy of $$\alpha-particle$$ is $$48\ MeV$$ and Q value of the reaction is $$50\ MeV$$. The mass number of the mother nucleus is: - (Assume that daughter nucleus is in ground state)

A
$$96$$

B
$$100$$

C
$$104$$

D
none of these

Solution

Let the reaction be:
$$M_1 \rightarrow \alpha + M_2 +Q$$

By conservation of energy:
$$Q = KE_1+KE_2$$
$$KE_1 = 48\ MeV$$
$$KE_2 = 50 -48 = 2\ MeV$$

Now conservation of momentum:
$$M_ {\alpha}V_{\alpha} = M_2V_2$$
Squaring both the sides and dividing by $$2$$.
Also, $$KE = \cfrac{1}{2}mv^2 = mp/2$$
$$m_{\alpha} =4$$

Substituting:
$$ 4 \times 48 = M_2 \times 2$$
$$M_2 = 96$$

Now by conservation of mass:
$$M_1 = M_{\alpha} + M_2$$
$$M_1 =100$$
Question 5
1 / -0

The binding energy of an electron in the ground state of $$He\ $$atom is $$E_0=24.6 eV.$$ The energy required to remove both the electrons from the atom is

A
$$24.6eV$$

B
$$79.0eV$$

C
$$54.4eV$$

D
None of these

Solution

Given, Energy required to remove electron in ground state from a neutral atom is $$24.6eV$$, which makes it $${He}^{+}$$ hydrogen like atom.
Energy required to remove electron from singly ionized helium is $$13.6\times{2}^{4}$$.
Therefore, Energy required to remove both the electrons is $$24.6+ 54.4= 79.0eV$$
Question 6
1 / -0

Find the binding energy of valence electron in the ground state of a $$Li$$ atom if the wavelength of the sharp series is known to be $$\lambda_{1}$$ $$= 813 nm$$ and the short wave cutoff wavelength $$\lambda_{2}$$, $$= 350 nm$$

A
$$3.54 eV$$

B
$$5.32 eV$$

C
$$1.5 eV$$

D
$$4.32 eV$$

Solution

For the first line of sharp series $$(3s\rightarrow 2p)$$ in a $$Li$$ atom
$$\displaystyle \frac{hc}{\lambda}=\frac{hR}{(3+\alpha _0)^2} + \frac{hR}{(2+\alpha_1)^2} $$ ...........(1)

For the short wave cut off wavelength of the same series
$$\displaystyle \frac{hc}{\lambda_{2}}= \frac{hR}{(2+\alpha_1)^2} $$ ...............(2)

Subtracting (1) from (2), we get:
$$(3+\alpha _0)^2$$$$=\displaystyle \frac{hR\lambda_{1}\lambda_{2}}{hc(\lambda_{1}-\lambda_{2})}=\frac{R\lambda_{1}\lambda_{2}}{2\pi c(\lambda_{1}-\lambda_{2})}$$

$$\displaystyle 3+\alpha _{0}=\sqrt\frac{R\lambda _{1}\lambda _{2}}{2\pi c(\lambda _{1}-\lambda _{2})}$$

In the ground state BE of electrons is:
$$=\displaystyle \dfrac{hR}{(2+\alpha _{0})^{2}}$$
$$\displaystyle =\dfrac{hR}{\left (\sqrt{\dfrac{R\lambda_{1}\lambda_{2}}{2\pi c(\lambda_{1}-\lambda_{2})}-1} \right )^{2}}=5.32 eV$$
Question 7
1 / -0

The above is a plot of binding energy per nucleon $$E_{b},$$ against the nuclear mass M; A, B,C, D, E, correspond to different nuclei. Consider four reactions :
(i) $$A+B\rightarrow C+\varepsilon $$
(ii) $$C\rightarrow A+B+\varepsilon $$
(iii) $$D+E\rightarrow F+\varepsilon $$
(iv) $$F\rightarrow D+E+\varepsilon ,$$
where $$\varepsilon$$ is the energy released. In which reactions $$\varepsilon $$ positive?

A
(i) and (iii)

B
(ii) and (iv)

C
(ii) and (iii)

D
(i) and (iv)

Solution

Lighter nuclei $$A$$ and $$B$$ can fuse to form a heavier nucleus C. In these fusion reactions, there is a mass defect giving rise to a release in energy. On the other hand, a heavy nucleus $$F$$ can be split into two nuclei $$D$$ and $$E$$ of moderate masses. In this fission reaction also, there is a mass defect which appears in the form of release of energy. Thus energy is released in the processes $$(i)$$ and $$(iv)$$.
Question 8
1 / -0

Masses of two isobars $$_{29}^{64}\textrm{Cu}$$ and $$_{30}^{64}\textrm{Zn}$$ are $$63.9298 amu$$ and $$63.9292 amu$$ respectively. It can be concluded from these data that

A
Both the isobars are stable

B
$$^{64}Zn$$ is radioactive, decaying to $$^{64}Cu$$ through $$\beta -$$ decay

C
$$^{64}Cu$$ is radioactive, decaying to $$^{64}Zn$$ through $$\lambda -$$ decay

D
$$^{64}Cu$$ is radioactive, decaying to $$^{64}Zn$$ through $$\beta -$$ decay

Solution

Since mass of the isobar has decreased, an electron has been ejected.
Thus $$^{64}_{29}Cu$$ decays to $$^{64}_{30}Zn$$ through the $$\beta - $$ decay as-
$$^{64}_{29}Cu\rightarrow^{64}_{30}Zn+^{0}_{-1}e$$
Question 9
1 / -0

Statement 1: Binding energy of $$_8O^{15}$$ is less than $$_7N^{14}$$
Statement 2 : Nuclear force is independent of the charge on the nucleons. For isobars, the difference in binding energies is mainly due to the difference in proton-proton coulombic repulsion

A
Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1

B
Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1

C
Statement 1 is True, Statement 2 is False

D
Statement 1 is False, Statement 2 is True.

Solution

Both nucleus have equal neutron number so they are isobars. The nuclear force is independent of charge. So the binding energy difference will come only the proton-proton coulomb repulsion.
Question 10
1 / -0

The following deuterium reactions and corresponding reaction energies are found to occur
$$^{14}N(d, p)^{15}N, Q=8.53 MeV$$
$$^{15}N(d, \alpha)^{13}C, Q=7.58 MeV$$
$$^{13}C(d, \alpha)^{11}B, Q=5.16 MeV$$
The rotation $$^{14}N(d, p)^{15}N$$ represents the reaction $$^{14}N+d\rightarrow ^{15}N+p$$
$$_2^4He=4.0026 amu, _1^2He=2.014 amu, _1^1H=1.0078 amu, n=1.0087 amu (1 amu=931 MeV)$$
The Q values of the reaction $$^{11}B(\alpha, n)^{14}N$$ is

A
$$0.5 eV$$

B
$$0.5 MeV$$

C
$$0.05 MeV$$

D
$$0.05 eV$$

Solution

The following reactions are as follows-
$$^{14}N (d,p)^{15}N : ^{14}N + d \rightarrow ^{15} N + p Q_1 = 8.53 MeV$$ ..............(i)
$$^{15}N (d,\alpha)^{13}C : ^{15}N + d \rightarrow ^{13} C + \alpha Q_2 = 7.58 MeV$$    ..............(ii)
$$^{13}C (d,\alpha)^{11}B : ^{13}C + d \rightarrow ^{11} B + \alpha Q_2 = 5.16 MeV$$     ...........(iii)
Adding (i), (ii) $ (iii), $$\implies$$ $$ ^{14}N + 3 d \rightarrow ^{11} B + 2\alpha +p $$ ...........(a)
$$Q$$ value for this reaction $$Q_a = Q_1 + Q_2 + Q_3 = 21.27 MeV$$
Now for reaction, $$3d \rightarrow p+ \alpha + n$$ ...........(b)
$$\therefore Q_b = [3\times 2.014 - 1.0078 + 4.0026 - 1.0087] \times 931 MeV = 21.32 MeV$$
Now $$(b) - (a) \implies $$ $$ ^{11}B + \alpha \rightarrow ^{14} N + n $$
$$\therefore$$ Q value for this reaction, $$Q = Q_b-Q_a = 21.32 - 21.27 = 0.05 MeV$$

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Nuclei Test - 60

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Nuclei Test - 60

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Question 1

Almost remaining 50% energy must be in form of kinetic energy of daughter nucleus.

Almost remaining 50% energy must be lost in form of heat.

Almost remaining 50% energy must be taken away by neutrino.

Almost remaining 50% energy must be taken away by antineutrino.

Solution

Question 2

increases continuously with mass number

decreases continuously with mass number

remains constant with mass number

first increases and then decreases with increase of mass number

Solution

Question 3

$$9.4 \times 10^9$$

$$9.4 \times 10^7$$

$$9.4 \times 10^{11}$$

$$9.4 \times 10^{13}$$

Solution

Question 4

$$96$$

$$100$$

$$104$$

none of these

Solution

Question 5

$$24.6eV$$

$$79.0eV$$

$$54.4eV$$

None of these

Solution

Question 6

$$3.54 eV$$

$$5.32 eV$$

$$1.5 eV$$

$$4.32 eV$$

Solution

Question 7

(i) and (iii)

(ii) and (iv)

(ii) and (iii)

(i) and (iv)

Solution

Question 8

Both the isobars are stable

$$^{64}Zn$$ is radioactive, decaying to $$^{64}Cu$$ through $$\beta -$$ decay

$$^{64}Cu$$ is radioactive, decaying to $$^{64}Zn$$ through $$\lambda -$$ decay

$$^{64}Cu$$ is radioactive, decaying to $$^{64}Zn$$ through $$\beta -$$ decay

Solution

Question 9

Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1

Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1

Statement 1 is True, Statement 2 is False

Statement 1 is False, Statement 2 is True.

Solution

Question 10

$$0.5 eV$$

$$0.5 MeV$$

$$0.05 MeV$$

$$0.05 eV$$

Solution

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