Nuclei Test

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Nuclei Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

Atomic weight of a radioactive element is $$M$$ gram/mol. Radioactivity of $$m$$ gram of its mass is :

A
$$N_{A}\ \lambda$$

B
$$\left(\dfrac {N_{A}m}{M}\right)\ \lambda$$

C
$$\left(\dfrac {N_{A}}{m}\right)\ \lambda$$

D
$$\left(\dfrac {N_{A}}{m}M\right)\ \lambda$$

Solution

$$\begin{array}{l}\text { Given, } \\\text { Atomic weight of element = Mgram / mol} \\\text { So, no of mole = } \frac{M}{M_{atomic}}\end{array}$$
$$\begin{array}{l}\text { Now no of atoms }=\text { mole } \times \text { Avagardo No. } \\\qquad N \quad=\frac{m}{M} \times N_{a}\end{array}$$
$$\begin{array}{l}\text { We knew, } \\\qquad \begin{aligned}\text { Radioactivity } &=\lambda \mathrm{N} \\R &=\frac{\lambda m}{M} \mathrm{Na} \\N &=\text { decay constant } \\N &=\text { no of atom } \\R &=(\frac{\mathrm{N_{a}}{m}}{M}) \lambda\end{aligned}\end{array}$$
Question 2
1 / -0

One mole of radium has an activity of 1/3.7 killo curie. Its decay constant will be

A
$$\dfrac{1}{6}\times 10^{-10}s^{-1}$$

B
$$ 10^{-10}s^{-1}$$

C
$$ 10^{-11}s^{-1}$$

D
$$ 10^{-8}s^{-1}$$

Solution

Number of mole of radium, $$n=1$$

The number of nuclei of radium, $$N={{N}_{a}}=avagdaro\,\,number=6.023\times {{10}^{23}}$$

Activity, $$A=\lambda N$$

$$ \because 1\,\,curie=3.7\times {{10}^{10}}\,decay/\sec $$

$$ \dfrac{1}{3.7}\,kilocurie=0.27\,kilocurie=9.99\times {{10}^{12}}\,dacay/\sec $$

So, $$\lambda =\dfrac{A}{N}=\dfrac{9.99\times {{10}^{12}}}{6.023\times {{10}^{23}}}=1.65\times {{10}^{-11}}\,$$
Question 3
1 / -0

A radioactive isotope has a half-life of $$2$$ years. How long will it approximately take the activity to reduce to $$3\%$$ of its initial value?

A
$$4.8\ Years$$

B
$$7\ Years$$

C
$$10\ Years$$

D
$$9.6\ Years$$

Solution

$$\begin{array}{l}=N_{0}\left(\frac{1}{2}\right)^{n} \\\text { Here, }\text {N is final activity. } \\\text { No is initial activity. }\\\text { Given: } \frac{N}{N_{0}}=\frac{3}{100}\\\therefore\quad\frac{3}{100}=\left(\frac{1}{2}\right)^{n}\\2^{n}=33.33 \\n \simeq 5\end{array}$$

$$\begin{array}{l}\therefore \text { Activity reduces to } 3 \% \text { of initial } \\\text { value of } 5 \text { half lives that is } 10\text { years. }\end{array}$$
Question 4
1 / -0

The nuclear radius of a certain nucleus is $$7.2 \,fm$$ and it has a charge of $$1.28 \times 10^{-17}C$$. The number of neutrons inside the nucleus is:

A
126

B
111

C
118

D
142

Solution

Radius of nuclear given by,
$$ R = ROA^{1/3} $$ where $$ Ro = 1.25\, fm $$
$$ \Rightarrow A = (\frac{R}{Ro})^{3} = (\frac{7.2}{1.25})^{3} = 191 $$ (atomic mass)
Now, no. of protons,
$$ Z = \frac{1.28\times 10^{-17}}{1.6\times 10^{-19}} = 80 $$
Thus, no.of neutrons nuclears:
$$ N = A-Z = 191-80 = 111 $$ (Ans)
Question 5
1 / -0

A particular hydrogen like atom has its ground state Binding Energy=122.4eV. It is in ground state. Then

A
atomic number is 3

B
an electron with 90 ev energy can interact with it and excite it

C
An electron of kinetic energy 91.8eV can be brought to almost rest by this atom.

D
An electron of KE nearly 2.6 eV may emerge from the atom electron of KE 125ev collides

Solution

$$\begin{array}{l}\text { Let Atomic number be "z" } \\\text { Binding energy }=z^{2}(13.6) \\\qquad \begin{array}{c}122.4=z^{2}(13 \cdot 6)\\z^{2}=9 \\z=3\end{array}\end{array}$$
Question 6
1 / -0

The kinetic energy of $$ \alpha $$ - particle emitted in the $$ \alpha $$ - decay of $$ { _{ 88 }{ Ra }^{ 226 } } $$ is [ given, mass number of Ra = 222 u ]

A
5.201 Me V

B
3.301 Me V

C
6.023 MeV

D
4.871 MeV

Solution

$$\begin{array}{l} _{ 88 }R{ a^{ 226 } }\to R_{ n }^{ 222 }+2H_{ e }^{ 4 } \\ Q=[226.02540-222.01750-4.00260]\times 931.5Mev \\ =0.0053\times 931.5=4.94Mev \\ K.E=\frac { { \left( { A-4 } \right) Q } }{ A } \\ =\frac { { 226-4 } }{ { 226 } } \times 4.94 \\ =4.85\, \, Mev \\ Q=4.94\, \, Mev, \\ K.E=4.85\, \, Mev \end{array}$$
Question 7
1 / -0

The active amount of radioactive substance left after one hour whose half life is $$20$$ minutes is?

A
$$\dfrac{1}{8}$$

B
$$\dfrac{1}{32}$$

C
$$\dfrac{1}{16}$$

D
$$\dfrac{1}{9}$$

Solution
Question 8
1 / -0

In a nuclear fusion reaction, two nuclei, $$A$$ & $$B$$, fuse to produce a nucleus $$C$$, releasing an atom of energy $$\Delta E$$ in the process. If the mass defects on the three nuclie are $$\Delta M_{A}, \Delta M_{B}, \Delta M_{C}$$ respectively, then which of the following relations holds? Here $$c$$ is the speed of light:-

A
$$\Delta M_{A}+\Delta M_{B}=\Delta M_{C}-\Delta E/c^{2}$$

B
$$\Delta M_{A}+\Delta M_{B}=\Delta M_{C}+\Delta E/c^{2}$$

C
$$\Delta M_{A}-\Delta M_{B}=\Delta M_{C}-\Delta E/c^{2}$$

D
$$\Delta M_{A}-\Delta M_{B}=\Delta M_{C}+\Delta E/c^{2}$$

Solution

$$\begin{array}{l}\text { Mass of photon is Difference in } \\\text { Mass defects of products and Reactants } \\\therefore\left[\Delta M_{C}\right]-\left[\Delta M_{A}+\Delta M_{B}\right]=\frac{\Delta E^{D}}{C^{2}} \\\triangle M_{A}+\Delta M_B=\Delta M_{C}-\frac{\Delta E}{C^{2}}\end{array}$$
Question 9
1 / -0

Assuming the energy released per fission of $$U^{235}$$ is 200 me V , the energy released in the fission of 1 kg of $$U^{235}$$ is energy

A
$$0.91\times10^{11}J$$

B
$$0.91\times10^{13}J$$

C
$$8.19\times10^{11}J$$

D
$$8.19\times10^{13}J$$

Solution

Given,

Energy released per atom, $$E=200\,MeV=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}=3.2\times {{10}^{-11}}\,J$$

Number of atom in $$1\,kg$$ uranium, $$n=\dfrac{m{{N}_{A}}}{M}=\dfrac{1000\times 6.023\times {{10}^{23}}}{235}=2.56\times {{10}^{24}}$$

Total Energy released = $$nE=2.56\times {{10}^{24}}\times 3.2\times {{10}^{-11}}=8.19\times {{10}^{13}}J$$

Hence, Total Energy released = $$8.19\times {{10}^{13}}J$$.
Question 10
1 / -0

The binding energy per nucleon of $$_{26}F^{56}$$ nucleus is

A
$$8.8\ MeV$$

B
$$88\ MeV$$

C
$$493\ MeV$$

D
$$413\ MeV$$

Solution

$$\begin{array}{l}\text { The binding energy per } \\\text { nucleon of } Fe \text { nucleus is } 8.8 \mathrm{MeV} \\\text { which is highest. }\end{array}$$

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Nuclei Test - 50

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Nuclei Test - 50

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Question 1

$$N_{A}\ \lambda$$

$$\left(\dfrac {N_{A}m}{M}\right)\ \lambda$$

$$\left(\dfrac {N_{A}}{m}\right)\ \lambda$$

$$\left(\dfrac {N_{A}}{m}M\right)\ \lambda$$

Solution

Question 2

$$\dfrac{1}{6}\times 10^{-10}s^{-1}$$

$$ 10^{-10}s^{-1}$$

$$ 10^{-11}s^{-1}$$

$$ 10^{-8}s^{-1}$$

Solution

Question 3

$$4.8\ Years$$

$$7\ Years$$

$$10\ Years$$

$$9.6\ Years$$

Solution

Question 4

126

111

118

142

Solution

Question 5

atomic number is 3

an electron with 90 ev energy can interact with it and excite it

An electron of kinetic energy 91.8eV can be brought to almost rest by this atom.

An electron of KE nearly 2.6 eV may emerge from the atom electron of KE 125ev collides

Solution

Question 6

5.201 Me V

3.301 Me V

6.023 MeV

4.871 MeV

Solution

Question 7

$$\dfrac{1}{8}$$

$$\dfrac{1}{32}$$

$$\dfrac{1}{16}$$

$$\dfrac{1}{9}$$

Solution

Question 8

$$\Delta M_{A}+\Delta M_{B}=\Delta M_{C}-\Delta E/c^{2}$$

$$\Delta M_{A}+\Delta M_{B}=\Delta M_{C}+\Delta E/c^{2}$$

$$\Delta M_{A}-\Delta M_{B}=\Delta M_{C}-\Delta E/c^{2}$$

$$\Delta M_{A}-\Delta M_{B}=\Delta M_{C}+\Delta E/c^{2}$$

Solution

Question 9

$$0.91\times10^{11}J$$

$$0.91\times10^{13}J$$

$$8.19\times10^{11}J$$

$$8.19\times10^{13}J$$

Solution

Question 10

$$8.8\ MeV$$

$$88\ MeV$$

$$493\ MeV$$

$$413\ MeV$$

Solution

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