Nuclei Test

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Nuclei Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

When $${ 10 }^{ 20 }$$ electrons are removed from a neutral metal plate through some process, the charge on it becomes ______

A
$$-1.6C$$

B
$$+16C$$

C
$$+1.6C$$

D
$${10}^{-19}C$$

Solution

$$Q={ 10 }^{ 20 }\times 1.6\times { 10 }^{ -19 }\\ \therefore Q=+16C$$
Question 2
1 / -0

A small quantity of solution containing $$^{24}Na$$ radio-nuclide (half life $$15$$ hours) of activity $$1.0$$ micro-curie is injected into the blood of a person. A sample of the blood of volume $$1\ cm^{3}$$ taken after $$5\ hours$$ shows an activity of $$296$$ disintegrations per minute. Determine the total volume of blood in the body of the person. Assume that the radio active solution mixes uniformly in the blood of the person.
($$1\ curie = 3.7\times 10^{10}$$ disintegrations per second).

A
$$5.95\ Litre$$.

B
$$5\ Litre$$.

C
$$55\ Litre$$.

D
$$7\ Litre$$.

Solution
Question 3
1 / -0

Statement 1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and
Statement II: For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decrease with increasing Z

A
Statement 1 is false, statement 2 is true

B
Statement 1 is false, statement 2 is true, statement 2 is correct explanation for statement 1

C
Statement 1 is false, statement 2 is true, statement 2 is not the correct explanation for statement 1

D
Statement 1 is true, statement 2 is false

Solution

$$ \begin{array}{l} \text { In nuclear fission or } \\ \text { fusion, energy is released. } \\ \text { Binding energy per nucleon } \\ \text { increase with increase } z \text { for } \\ \text { heavy nuclei and decrease with } \\ \text { decrease } z \text { for light nuclei. } \\ \text { Both are true } \\ \text { and correct explanation. } \end{array} $$
Question 4
1 / -0

A nucleus $$X$$, initially at rest, undergoes alpha-decay according to the equation $$_{92}^{A}X \rightarrow _{Z}^{228}Y + \alpha$$.
The alpha particle produced in the above process is found to move in a circular track of radius $$0.11\ m$$ in a uniform magnetic field of $$3\ T$$. Find the energy $$(ln\ MeV)$$ released during the process and the binding energy of the parent nucleus $$X$$.
Given that $$m(Y) = 228.03\ u; m(_{0}^{1}n) = 1.009u . m (_{2}^{4}He) = 4.003u; m(_{1}^{1} H) = 1.009 u$$.

A
$$5.34\ MeV, 1823\ MeV$$.

B
$$7\ MeV, 1823\ MeV$$.

C
$$9\ MeV, 1458\ MeV$$.

D
$$6\ MeV, 23\ MeV$$.

Solution
Question 5
1 / -0

Binding Energy per nucleon of a fixed nucleus $$X^4$$ is $$6\ MeV.$$ It absorbs a neutron moving with $$KE = 2\ MeV,$$ and converts into $$Y$$ at ground state, emitting a photon of energy $$1\ MeV.$$ The Binding Energy per nucleon of $$Y$$ ($$in Mev$$) is:

A
$$\dfrac{(6A + 1)}{(A + 1)}$$

B
$$\dfrac{(6A - 1)}{(A + 1)}$$

C
$$7$$

D
$$\dfrac{7}{6}$$

Solution

$$\begin{array}{l}\text { Given-* Binding Energy per nucleon of } X^{A}=6 \text { Mev } \\\Rightarrow \text { Binding Energy } B E \text { of } X^{A}=6 A \text { MeV } \\\text { we have - }\\\qquad\begin{array}{c}X^{A}+_1^1{} n \longrightarrow Y^{A+1}+\text { Photon (l Mev) } \\ \end{array}\end{array}$$

$$\begin{aligned}\text { Initial Energy } &=B E \text { of }X^{A}+\text { Energy of neutron } \\&=(6 A+2) \mathrm{MeV} \\\text { Final Energy }&=B E \text { of } Y^{A+1}+\text { Energy of photon } \\&=B E_{Y}+1\mathrm{MeV}\end{aligned}$$

$$\begin{array}{l}\text { By energy conservation- }\\\qquad\begin{aligned}(6 A+2) \mathrm{meV}=B E_{Y}+\text { 1 MeV }\\\Rightarrow B E \text { of } Y=(6 A+1) \text { MeV } \\\Rightarrow \text { Binding Energy per nucleon }=\\B E=\frac{6 A+1}{A+1} \text { Mev }\end{aligned} \\\text { Hence, opt (A) is correct. }\end{array}$$
Question 6
1 / -0

If the binding energy of the electron in a hydrogen atom is $$13.6\ eV$$, the energy required to remove the electron from the first excited state of $${Li}^{2+}$$ is:

A
$$30.6\ eV$$

B
$$13.6\ eV$$

C
$$3.4\ eV$$

D
$$122.4\ eV$$

Solution

$$E$$ of $${ Li }^{ ++ }$$ at $$n=2=\cfrac { \left( 13.6 \right) { 3 }^{ 2 } }{ { 2 }^{ 2 } } =30.6eV$$
Question 7
1 / -0

In $$Q-26$$, find the initial activity of sample.

A
$$6.79\ \times 10^{16}\ disintegration/sec$$

B
$$5.79\ \times 10^{15}\ disintegration/sec$$

C
$$10\ \times 10^{15}\ disintegration/sec$$

D
none of these.

Solution
Question 8
1 / -0

A nuclear transformation is denoted by $$X\left( n,\alpha \right) \ _{ 3 }^{ 7 }{ Li }$$. Which of the following is the nucleus of element $$X$$?

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

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

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

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

Solution

$$X+n=_{ 2 }^{ 4 }{ He }+_{ 3 }^{ 7 }{ Li }\quad $$
So. from conservation of mass no.
$$X$$ must be $$_{ 5 }^{ 10 }{ B }$$
Question 9
1 / -0

In the nuclear fusion reaction
$$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 3 }{ H }\rightarrow _{ 2 }^{ 4 }{ He }+n\quad $$ given that the repulsive potential energy between the two nuclei is $$-7.7\times { 10 }^{ -14 }J$$, the temperature at which the gases must be heated to initiate the reaction is nearly (Boltzmann's constant $$k=1.38\times { 10 }^{ -23 }J/K$$)

A
$${ 10 }^{ 7 }K\quad $$

B
$${ 10 }^{ 5 }K$$

C
$${ 10 }^{ 3 }K$$

D
$${ 10 }^{ 9 }K$$

Solution

From conservation of energy
$$\quad 2\cfrac { 1 }{ 2 } m{ v }_{ }^{ 2 }=7.7\times { 10 }^{ -14 }$$
$$\therefore v=\sqrt { \cfrac { 7.7\times { 10 }^{ -14 } }{ m } } =\sqrt { \cfrac { 3kT }{ m } } \quad \quad \therefore T\approx { 10 }^{ 9 }K$$
Question 10
1 / -0

If $$I$$ excitation energy for the $$H-$$ like (hypothetical) sample is $$24\ eV$$, then binding energy in $$III$$ excired state is:

A
$$2\ eV$$

B
$$3\ eV$$

C
$$4\ eV$$

D
$$5\ eV$$

Solution

$$\text{First(I) excitation energy means energy difference between first(ground) state and second state. }$$
If ground state energy is $$E_1$$ then for $$n=2$$ the energy becomes $$E_1/n^2$$
Note- As the energy shown above is due to attractive force so it is $$negative$$ in magnitude.
So $$E_2=E_1/2^2=E_1/4$$
difference is $$E_1-E_2=3E_1/4$$
Excitation energy will be $$negative $$ of the energy difference.
Given that $$-3E_1/4=24eV$$ so $$E_1=-32eV$$

$$\text{Third(III) excited state means fourth normal state i.e n=4 }$$
Energy of $$4^{th} state $$ is $$E_1/4^2=E_1/16=-32/16=-2eV$$

Binding energy will be positive in nature so answer is $$2eV$$

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Re-Attempt Weekly Quiz Competition

Nuclei Test - 47

Result

Nuclei Test - 47

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SHARING IS CARING

Question 1

$$-1.6C$$

$$+16C$$

$$+1.6C$$

$${10}^{-19}C$$

Solution

Question 2

$$5.95\ Litre$$.

$$5\ Litre$$.

$$55\ Litre$$.

$$7\ Litre$$.

Solution

Question 3

Statement 1 is false, statement 2 is true

Statement 1 is false, statement 2 is true, statement 2 is correct explanation for statement 1

Statement 1 is false, statement 2 is true, statement 2 is not the correct explanation for statement 1

Statement 1 is true, statement 2 is false

Solution

Question 4

$$5.34\ MeV, 1823\ MeV$$.

$$7\ MeV, 1823\ MeV$$.

$$9\ MeV, 1458\ MeV$$.

$$6\ MeV, 23\ MeV$$.

Solution

Question 5

$$\dfrac{(6A + 1)}{(A + 1)}$$

$$\dfrac{(6A - 1)}{(A + 1)}$$

$$7$$

$$\dfrac{7}{6}$$

Solution

Question 6

$$30.6\ eV$$

$$13.6\ eV$$

$$3.4\ eV$$

$$122.4\ eV$$

Solution

Question 7

$$6.79\ \times 10^{16}\ disintegration/sec$$

$$5.79\ \times 10^{15}\ disintegration/sec$$

$$10\ \times 10^{15}\ disintegration/sec$$

none of these.

Solution

Question 8

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

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

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

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

Solution

Question 9

$${ 10 }^{ 7 }K\quad $$

$${ 10 }^{ 5 }K$$

$${ 10 }^{ 3 }K$$

$${ 10 }^{ 9 }K$$

Solution

Question 10

$$2\ eV$$

$$3\ eV$$

$$4\ eV$$

$$5\ eV$$

Solution

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