Physics Part Ii Chapter 13 Nuclei

(a) Electromagnetic radiation used for water purification: U.V (Ultra Violet) rays.

(b)  Electromagnetic radiation used for eye surgery: Infrared Radiation (Laser light).

Nuclear Fission:

When a heavy nucleus (A > 235 say) breaks into two lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e, nucleons get more tightly bound. This implies that energy would be released in nuclear fission.

Nuclear Fusion:

When two very light nuclei join to form a heavy nucleus, the binding is energy per nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion.

Example:${}_{1}H^2 + {}_{1}H^2 \underset{ }{\to } {}_{2}H{e}^4 + 26 Mev$

BE Graph:

A₀  = 100

A_t = 3.125

A_t = A₀.e^-λt

$$\begin{gathered} 3.125 = 100 {\mathrm{e}}^{\frac{-0.693}{10}\mathrm{t}} \\ \frac{3.125}{100} = {\mathrm{e}}^{-0.0693 \mathrm{t}} \\ \frac{100}{3.125} = {\mathrm{e}}^{0.0693 \mathrm{t}} \\ {\log }_{\mathrm{e}} \frac{100}{3.125} = 0.0693 \mathrm{t} \\ {\mathrm{t}}_2 = \frac{2.303 {\log }_{10}^{12}}{0.0693} \\ \mathrm{t} = 50.1 \mathrm{years} \end{gathered}$$

A.

(i) and (iv)

Binding energy per nucleon of each product is less than that of each reactant.

D.

5→ 4

IR corresponds to least value of  i.e. from Paschen, Bracket and Pfund series. Thus the transition corresponds to 5 → 4

D.

Statement – 1 is true, Statement – 2 is False.

B.

(M₀ – ⁸M_P – ⁹M_N) C²

B.

there is no change in the proton number and the neutron number

B.

Y will decay faster than X

D.

gamma photons

D.

the biological effect of radiation

C.

17.28 MeV

E_P = (8 × 7.06 – 7 × 5.60) MeV
= 17.28 MeV

D.

12 mm

B.

increase by a factor of 4

C.

5 minutes

A.

6 fermi

B.

C.

3F/8

B.

1:3^1/2

Law of conservation of momentum gives
m₁ v₁ = m₂ v₂

C.

23.6 MeV

As given,
The binding energy per nucleon of a deuteron (₁H² ) = 1.1 MeV
∴ Total binding energy = 2× 1.1 = 2.2 MeV
The binding energy per nucleon of helium (₂He⁴ ) = 7 MeV
∴ Total binding energy = 4× 7 = 28 MeV
Hence, energy released in the above process = 28 - 2× 2.2 = 28 - 4.4 = 23.6 MeV

C.

10⁻¹² cm

At closest approach, all the kinetic energy of the α-particle will converted into the potential energy of the system, K.E. = P.E.

B.

1/7λ

A.

20

Number of nuclei remaining,

N = 600-450 = 150 after time 't'

$$\begin{gathered} \frac{N}{N_0} = \frac{150}{600} = \frac{1}{4} \\ \frac{150}{600} = {\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}} \\ {\left(\frac{1}{2}\right)}^2 = {\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}} \\ t = 2t_{1/2} = 2 x 10 \\ = 20 minute \end{gathered}$$

A.

${}_{83}\mathrm{X}^{214}$

Due to the emission of 3α particles, the mass number is reduced by 12 while the atomic number is decreased by 6. Due to the emission of β particle, the atomic number is increased by one while there is no change in mass number. Now, resulting mass will be decreased by 12 and atomic number is decreased by 5. So, isotopes X is ₈₃X²¹⁴

B.

${\left(\frac{\mathrm{NR}}{\mathrm{r}}\right)}^{1/2}$

As each is containing n lamps, hence,

The resistance of each arc = nr

Number of arcs = N/n

Total  resistance S is given by

$$\begin{gathered} \frac{1}{\mathrm{S}} = \mathrm{\Sigma }\frac{1}{\mathrm{nr}} = \frac{\mathrm{N}}{\mathrm{n}}\left(\frac{1}{\mathrm{nr}}\right) \\ \mathrm{S} = \frac{{\mathrm{n}}^2\mathrm{r}}{\mathrm{N}} \\ \therefore \mathrm{Total} \mathrm{resistance} = \mathrm{R} + \mathrm{S} + \left({\mathrm{n}}^2\mathrm{r}\right)\mathrm{N} \end{gathered}$$

If E is the emf of the machine, current entering the arcs is E/(R+S) and in each arc is nE/(R+S)N.

Hence, the current passing through each lamp.

$\mathrm{I} = \frac{\mathrm{nE}}{\mathrm{N}(\mathrm{R} +{\mathrm{n}}^2\mathrm{r}/\mathrm{N})} = \frac{\mathrm{E}}{\mathrm{N}}{\left[\frac{\mathrm{R}}{\mathrm{n}} + \frac{\mathrm{nr}}{\mathrm{N}}\right]}^{-1}$

Now heat produced per second in the lamps is 

H = NI²

since, light emitted is proportional to H2 therefore, the light produced is maximum when H² and hence H is maximum or $\left[\frac{\mathrm{R}}{\mathrm{n}} + \frac{{\displaystyle \mathrm{nr}}}{\mathrm{N}}\right]$ is minimum. Hence, we can write,

$\frac{\mathrm{R}}{\mathrm{n}} + \frac{\mathrm{nr}}{\mathrm{N}} = \left[{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)}^{1/2} - {\left(\frac{\mathrm{nr}}{\mathrm{N}}\right)}^{1/2}\right] + 2{\left(\frac{\mathrm{Rr}}{\mathrm{N}}\right)}^{1/2}$

This is minimum when $\sqrt{\mathrm{R}/\mathrm{n}} - \sqrt{\mathrm{nr}/\mathrm{N}} = 0$ or very small or n is closely equal to (NR/r)^1/2

C.

lead

Let Mo be initial mass of Rn -220

Number of half-lives of Rn in 5 minutes = 5min/55s

$$\begin{gathered} = \frac{5 \mathrm{x} 60}{55} \approx 5.5 \\ \therefore \mathrm{Mass} \mathrm{of} \mathrm{Rn} - 220 \left(\mathrm{left}\right) = {\left(\frac{1}{2}\right)}^{5.5} \\ {\mathrm{M}}_{\mathrm{o}} =\frac{{\mathrm{M}}_0}{5} \end{gathered}$$

Therefore, the mass converted into Po - 216 is (44/45)

Mo, number of half-lives of Po- 216 in 5 minutes

$$\begin{gathered} = \frac{5 \min }{0.66} = \frac{300}{0.66} \approx 455 \\ \therefore \mathrm{Mass} \mathrm{of} \mathrm{Po} - 216 \mathrm{left} \\ = {\left(\frac{1}{2}\right)}^{455} \mathrm{x} \frac{44}{45} {\mathrm{M}}_0 \stackrel{ }{\to } 0 \end{gathered}$$

This means that Po - 216 formed will soon decay to lead. Hence, the lead will have maximum mass nearly (44/45)M₀

B.

$\mathrm{BE} = [{\mathrm{Zm}}_{\mathrm{p}} + (\mathrm{A} -\mathrm{Z}){\mathrm{m}}_{\mathrm{n}} - \mathrm{m}(\mathrm{A},\mathrm{Z}\left)\right]{\mathrm{c}}^2$

In the case of the formation of the nucleus, the evolution of energy equals the binding energy of the nucleus takes place due to the disappearance of a fraction of total mass. If the quantity of mass disappearing is Δm, then the binding energy is

BE = Δmc²

From the above points, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons. Thus, 

Δm = Zm_p - Nm_n - m(A,Z)

Where m (A, Z) is the mass of the atom of mass number A atomic number Z. Hence, the binding energy of a nucleus is

BE = [Zm_p + Nm_n - m(A,Z)]c²

BE = [Zm_p + (A-Z) mn - m(A,Z)] c²

C.

Y will decay at a faster rate than X

The half-life of x - mean life of y

Let λx and λy be decay constants of x and y, then condition (1) implies

$$\begin{gathered} \frac{0.693}{{\mathrm{\lambda }}_{\mathrm{x}}} = \frac{1}{{\mathrm{\lambda }}_{\mathrm{y}}} \\ {\mathrm{\lambda }}_{\mathrm{y}} = \frac{{\mathrm{\lambda }}_{\mathrm{x}}}{0.693} \\ \Rightarrow {\mathrm{\lambda }}_{\mathrm{y}} > {\mathrm{\lambda }}_{\mathrm{x}} \end{gathered}$$

As rate of decay αλ; hence decay of y will be faster than that of x.

D.

both the trajectories will be equally curved.

$$\begin{gathered} \mathrm{K} = \frac{1}{2}{\mathrm{mV}}^2, \\ \mathrm{In} \mathrm{electric} \mathrm{field}, \\ \mathrm{Y} = \frac{{\mathrm{qEx}}^2}{2{\mathrm{mV}}^2} \\ = \frac{{\mathrm{qEx}}^2}{4\mathrm{K}} \end{gathered}$$

The curvature of the trajectory is proportional to qE/K. Since q, E and K are the same for both particles, hence the trajectories will have the same curvature.

B.

3α, 4β

$$\begin{gathered} {}_{84}\mathrm{A}^{222} \to {}_{84}\mathrm{B}^{210} \\ \mathrm{Decrease} \mathrm{in} \mathrm{mass} \mathrm{number} = 222 -210 \\ = 12 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{\alpha } \mathrm{particles} \mathrm{emitted} \\ =\frac{12}{4} \\ = 3 \\ {}_{86}\mathrm{A}^{222} \to {}_{80}\mathrm{X}^{210} \to {}_{84}\mathrm{B}^{210} \end{gathered}$$

Increase in atomic number = 84 - 80 = 4

∴ Number of β particless emitted = 4

C.

nuclear process does not depend on external factors

Radioactivity is a natural process which does not depend upon external factors.

B.

c - a - b

The energy released per nuclear reaction is the resultant binding energy.

$$\begin{gathered} \mathrm{Binding} \mathrm{energy} \mathrm{of} \left( {}_1{}^2\mathrm{H} + {}_1{}^3\mathrm{H} \right) = \mathrm{a} + \mathrm{b} \\ \mathrm{Binding} \mathrm{energy} \mathrm{of} {}_2{}^4\mathrm{He} = \mathrm{c} \end{gathered}$$

In a nuclear reaction, the resultant nucleus is more stable than the reactants. Hence, the binding energy of  ${}_2{}^4\mathrm{He}$  will be more than that of  $\left({}_1{}^2\mathrm{H} + {}_1{}^3\mathrm{H}\right)$

Thus, the energy released per nucleon

   = resultant binding energy

  = c - ( a+ b)

 = c - a - b

B.

decreases with mass number at high mass numbers

The binding energy per nucleon for the middle nuclides (from A= 20 to A= 56) is maximum. Hence, these are more stable. As the mass number increases, the binding energy per nucleon gradually decreases and ultimately binding energy per nucleon of heavy nuclides (such as uranium etc.) is comparatively low. Hence, these nuclides are relatively unstable so they can be finished easily.

A.

less than 1

In fission process, when a parent nucleus breaks into daughter products, then some mass is lost in the form of energy. Thus, mass of fission products < mass of parent nucleus

$\Rightarrow \underset{\mathrm{Mass} \mathrm{of} \mathrm{friend} \mathrm{nucleus}}{\overset{\mathrm{Mass} \mathrm{of} \mathrm{fission} \mathrm{products}}{\to }}$< 1

B.

the electrons produced as a result of the decay of neutrons inside the nucleus

(1) Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a β-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.
(2) In negative β-decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

D.

$\frac{1}{2\mathrm{\lambda }}$

Number of nuclei remained after time t can be written as
N = N_o e^-λt
where N_o is initial number of nuclei of both the substances.

N₁ = N_o e^-5λt                   ....... (i)

and N₂ = N_o e^-λt               ...... (ii)

Dividing Eqⁿ (i) by Eqⁿ (ii), we obtain

$$\begin{gathered} \frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}={\mathrm{e}}^{ \left(-5 \mathrm{\lambda } + \mathrm{\lambda }\right)\mathrm{t}} \\ = {\mathrm{e}}^{-4\mathrm{\lambda t}} \\ = \frac{1}{{\mathrm{e}}^{4\mathrm{\lambda t}}} \end{gathered}$$

But we have given

$$\begin{gathered} \frac{{\mathrm{N}}_1}{{\mathrm{N}}_2} = {\left(\frac{1}{\mathrm{e}}\right)}^2 \\ = \frac{1}{{\mathrm{e}}^2} \\ \frac{1}{{\mathrm{e}}^2} = \frac{1}{{\mathrm{e}}^{4\mathrm{\lambda t}}} \end{gathered}$$

Comparing the powers we get

 2 = 4λt

$$\begin{gathered} \mathrm{t} = \frac{2}{4\mathrm{\lambda }} \\ \mathrm{t} = \frac{1}{2\mathrm{\lambda }} \end{gathered}$$

B.

( M_o - 8 M_P - 9 M_c ) c²

Binding energy:- If a certain number of protons and neurons are brought together to form a nucleus to form a nucleus of a certain charge and mass, an energy E_b will be released in the process. The energy E_b is called the binding energy of the nucleus.

BE = ( M_nucleus - M_nucleons ) c²

BE = (M_o - 8M_P - 9 M_n ) c²

B.

fusion

The working of hydrogen bomb is based on nuclear fusion. But for the process of nuclear fusion, a very high temperature (≈ 10⁷ K) and a very high pressure is required, hence to construct a hydrogen bomb, first atom bomb is constructed.

The atom bomb is then covered from all sides by an enclosure of a compound of heavy hydrogen (deuteron) such as lithium hydride. First the atom bomb is exploded which produces such a high temperature and pressure that deuteron nuclei get fused and a huge amount of energy is released.

A.

1

The multiplication factor (k) is an important reactor parameter and is the ratio of number of neutrons present at the beginning of a particular generation to the number present at the beginning of the next generation. It is a measure of the growth rate of the neutrons in the reactor. For k =l, the operation of the reactor is said to be critical.

 Note: If k becomes greater than one, the reaction rate and the reactor power increase exponentially. Unless the factor k is brought down very close to unity, the reactor will become supercritical and can even explode.

C.

1 : 4

Total number of nuclei remained after n half- lives is 

             N = N_o ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

Total time given = 80 min

Number of half-lives of A, n_A = $\frac{80 \min }{20 \min }$

                                           = 4

Number of half-lives of B, n_B = $\frac{80 \min }{40 \min }$

                                           = 2

Number of nuclei remains undecayed

                    $\mathrm{N} ={\mathrm{N}}_{\mathrm{o}} {\left(\frac{1}{2}\right)}^{\mathrm{n}}$

Where N_o is initial number of nuclei.

            $\therefore \frac{{\mathrm{N}}_{\mathrm{A}}}{{\mathrm{N}}_{\mathrm{B}}} = \frac{{\displaystyle {\left(\frac{1}{2}\right)}^{{\mathrm{n}}_{\mathrm{A}}}}}{{\displaystyle {\left(\frac{1}{2}\right)}^{{\mathrm{n}}_{\mathrm{B}}}}}$

            $$\begin{gathered} \Rightarrow \frac{ {\mathrm{N}}_{\mathrm{A}}}{{\mathrm{N}}_{\mathrm{B}}} = \frac{{\left({\displaystyle \frac{1}{2}}\right)}^4}{{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =\frac{{\displaystyle \frac{1}{16}}}{{\displaystyle \frac{1}{4}}} \end{gathered}$$

               $\Rightarrow \frac{{\mathrm{N}}_{\mathrm{A}}}{{\mathrm{N}}_{\mathrm{B}}} = \frac{1}{4}$

C.

3.125 g

   t_1/2 = 4 hour

     $\mathrm{n} = \frac{\mathrm{T}}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$      

         = $\frac{24}{4}$

      n = 6

  M = M_o ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

     = $200 {\left(\frac{1}{2}\right)}^6$

  M = 3.125 g

B.

₈₂Pb²⁰⁶

After one $\mathrm{\alpha }$-emission, the daughter nucleus reduces in mass number by 4 unit and in atomic number by 2 unit. In [β-emission the atomic number of daughter nucleus increases by 1 unit.

The reaction can be written as

${}_{92}\mathrm{U}^{238}\stackrel{-8\mathrm{\alpha }}{\to } {}_{ 76}\mathrm{X}^{206} \stackrel{-6\mathrm{\beta }}{\to } {}_{82}\mathrm{Y}^{206}$

Thus the resulting nucleus is  ₈₂Y²⁰⁶, i.e  ₈₂Pb²⁰⁶

C.

4.8 x10⁷ C/kg

Specific charge on proton = ${\left(\frac{\mathrm{e}}{\mathrm{m}}\right)}_{\mathrm{P}}$

                                 = 9.6 x 10⁷ C/kg

Let charge of proton be +e, then charge of alpha particle be +2e. Similarly, let mass of proton be m, then mass of alpha particle will be 4m

Specific charge = charge/ mass of the substance

Specific charge on $\mathrm{\alpha }$-particle,

${\left(\frac{\mathrm{q}}{\mathrm{m}}\right)}_{\mathrm{\alpha }}$ = $\frac{2\mathrm{e}}{4\mathrm{m}}$

          = $\frac{1}{2}{\left(\frac{\mathrm{e}}{\mathrm{m}}\right)}_{\mathrm{P}}$

          = $\frac{1}{2} \times 9.6 \times {10}^7$

${\left(\frac{\mathrm{q}}{\mathrm{m}}\right)}_{\mathrm{\alpha }} = 4.8 \times {10}^7 \raisebox{1ex}{$\mathrm{C}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Kg}$}\right.$

A.

(1/27) of the original value

   R =R_o ${\mathrm{e}}^{-\mathrm{\lambda t}}$

⇒ $\frac{1}{3} =$e^{-λ × 3}

        = e^-3λ              ...(i)

Again 

    $\frac{\mathrm{R}\text{'}}{{\mathrm{R}}_{\mathrm{o}}}$= e^{-λ × 9}

          = e^-9λ

           =( e^3λ )³

           = ${\left(\frac{1}{3}\right)}^3$

            = $\frac{1}{27}$

     $\frac{\mathrm{R}\text{'}}{ {\mathrm{R}}_{\mathrm{o}} }= \frac{{\mathrm{R}}_{\mathrm{o}}}{27}$

Hence, in 9 days activity will become $\left(\frac{1}{27}\right)$ of the original value.

D.

$\frac{4}{5}$

Nucleus of mass number A has a radius   

      $\mathrm{R} = {\mathrm{R}}_{\mathrm{o}} {\left(\mathrm{A}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Atomic mass number A₁ =64

              ${\left(64 \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ = 4

Atomic mass number A₂ = 125

               ${\left(125\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right. } = 5$

∴        $\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2} = {\left(\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

              = ${\left[ {\left(\frac{4}{5}\right)}^{3 }\right]}^{ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

          $\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2} = \frac{4}{5}$

A.

9hcR

Ionization energy is the energy necessary to remove an electron from the neutral atom. 

Ionization energy =  RchZ² 

                         Z = 3 for Li²⁺

∴  Iionization energy = ( 3 )² Rch

                                = 9Rch

A.

1

     ${}_{83 }\mathrm{X}{ }^{238} \to {}_{83 }\mathrm{Y}^{ 222}$

α - particles are emitted = $\frac{238 - 222}{4}$

                                    = 4

The atomic number is decreased 

                                 = 90 - 4 × 2 

                                  = 82

As the atomic number of ${}_{83}\mathrm{Y}^{ 222}$, so the atomic number is increased by 1, therefore, one β-particle is emitted.

C.

${}_{26}{}^{56}\mathrm{Fe}$

We know that binding energy per nucleon increases with atomic number binding energy for iron is maximum, after that it decreases.

C.

3.125 g

   half-life of radio isotope = 4 h

   Initial mass M_o = 200 g   

            M = M_o ${\left(\frac{1}{2}\right)}^{ \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{t}}_{1/2}$}\right.}$

     ∴          M = 200 ${\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$24$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

                    = $200 {\left(\frac{1}{2}\right)}^6$

                    = $\frac{200}{64}$

              M =  3.125 g

B.

3.125%

Remaining quantity

   N = N_o ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

        = N_o ${\left(\frac{1}{2}\right)}^5$

   N = $\frac{{\mathrm{N}}_{\mathrm{o}}}{32}$

In percentage,

  N =  $\frac{{\mathrm{N}}_{\mathrm{o}}}{30 \times {\mathrm{N}}_{\mathrm{o}}}\times 100$

  N = 3.125%

C.

3.9 days

The half-life and the decay constant are related as

              t₁₂ = $\frac{\ln 2}{\mathrm{\lambda }}$ = $\frac{0.693}{\mathrm{\lambda }}$

⇒              $\mathrm{\lambda } = \frac{0.693}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$ 1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$         

                     = $\frac{0.693}{2.7 \mathrm{day}}$

                     = $\frac{0.693}{2.7 \times 24 \times 3600 \mathrm{s}}$

since 2.7 day = 2.7 × 24 hours × 3600s 

             λ = 0.0000029 s^-1

             λ = 2.9 × 10^-6 s^-1

Now average life

             λ = $\frac{1}{\mathrm{\lambda }}$ = 3.9 days

D.

( 8 M_p + 9 M_n - M_O)

Given:- ₈O¹⁷ ⇒ Mass number of oxygen = 17                

The binding energy is given as

       BE = Δm c²

Where Δm = mass defect

         c = speed of light

      Δm = [ Z M_p + (A - Z ) M_n ] - M

        Z - atomic number = number of protons

      N - neutron number  = number of neutrons

     A - mass number = Z + N

                               = Total number of protons + total number of neutrons

      Δm = [ 8 M_p + ( 17 - 8 )M_n - M_O

⇒    BE = ( 8 M_p + 9 M_n - M_O ) c²

B.

1 : 3^1/2

D.

1 : 2^1/2

Given:- A nucleus disintegrate into two parts

         Their velocities ratio v₁ : v₂ = 2 : 1

According to the law of conservation of momentum, we get

               m₁ v₁ = m₂ v₂

⇒               $\left(\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}\right) = \left(\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}\right)$

But        m = $\frac{4}{3} \mathrm{\pi } {\mathrm{r}}^3\mathrm{\rho }$

               m ∝ r³

⇒           $\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2} = \frac{{\mathrm{r}}_1^3}{{\mathrm{r}}_2^3}$

∴             $\frac{{\mathrm{r}}_1^3}{{\mathrm{r}}_2^3} = \frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}$

⇒             $\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2} = {\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

∴              r₁ : r₂ = 1 : (2)^1/3  

B.

45 yr

Given:- T_1/2 = 15 yr

           N₀ = 12 g

          N = 1.5 g

∴           N = N₀ e^-λt

Where N is the number of radioactive nuclei at time t       

λ = radioactive decay constan or disintegration constant

⇒         e^-λt = $\frac{\mathrm{N}}{{\mathrm{N}}_0}$

Take log on both side

⇒          $- \mathrm{\lambda } \mathrm{t} = \ln \left(\frac{\mathrm{N}}{{\mathrm{N}}_0}\right)$

⇒         t = $\frac{1}{\mathrm{\lambda }} \ln \left(\frac{{\mathrm{N}}_0}{\mathrm{N}}\right)$

⇒            = $\frac{{\mathrm{T}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{0.693}$ $\ln \left(\frac{{\mathrm{N}}_0}{\mathrm{N}}\right)$                        

Since      ${\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} = \frac{\ln 2}{\mathrm{\lambda }} = \frac{0.693}{\mathrm{\lambda }}$                             

⇒            = $\frac{15}{0.693} \ln \left(\frac{{\mathrm{N}}_0}{\mathrm{N}}\right)$

⇒         e^-λt = $\frac{15}{0.93} \times \ln \left(\frac{12}{1.5}\right)$

⇒          e^-λt = 45 yr

A.

0.87 × 10⁷ s

Given:-

  i = 10^-4 A

  t = 1h = 3600 s

We know that

  q = it

      = 10^-4 × 1 × 3600

  q = 0.36 C

∴ Number of protons = $\frac{\mathrm{q}}{\mathrm{e}}$

                              = $\frac{0.36}{1.6 \times {10}^{-19}}$                [ e = 1.6 × 10^-19 C ]

                              = 2.25 × 10⁸

Number of ⁷Be nuclei produced

                               = $\frac{2.25 \times {10}^8}{1000}$

                                 = 2.25 × 10⁸

Activity of ⁷Be = 1.8 × 10⁸ disintegration/s

Acivity A = λ N

            = $\frac{\ln 2}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ N                       $\left[\because \mathrm{\lambda } = \frac{\ln 2}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right]$

       1.8 × 10⁶ = $\frac{0.693}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$ 1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \times 2.25 \times {10}^{15}$

       t_1/2 = $\frac{0.693 \times 2.25 \times {10}^{15}}{1.8 \times {10}^8}$

        t_1/2 = 0.87 × 10⁷ s

B.

$\frac{1}{3 \mathrm{\lambda }}$

The ratio of number of nuclei of X₁ to that of X₂  after time t,

∴           $\frac{{\mathrm{N}}_1 }{{\mathrm{N}}_2} = \frac{1}{\mathrm{e}}$  

∴         $\frac{{\mathrm{N}}_0 {\mathrm{e}}^{-6\mathrm{\lambda t}}}{{\mathrm{N}}_0 {\mathrm{e}}^{-3\mathrm{\lambda t}}} = \frac{1}{\mathrm{e}}$                          ( $\because$ N = N₀ e^-λt )

         $\frac{1}{{\mathrm{e}}^{ 3\mathrm{\lambda t}}} = \frac{1}{\mathrm{e}}$

⇒         3λt = 1

∴             t = $\frac{1}{3 \mathrm{\lambda }}$

D.

$\frac{48}{169}$

Let m and M be the masses of neutron and carbon nucleus (at rest) respectively. 

If u and v are the velocity of neutron before and after collision, then

           K_i = $\frac{1}{2}$mu²

and    K_f = $\frac{1}{2}$ mv²

But   v = $\frac{\left( \mathrm{m} - \mathrm{M} \right) \mathrm{u}}{\mathrm{m} + \mathrm{M}}$

∴     K_f = $\frac{1}{2} \mathrm{m} {\left(\frac{\mathrm{m} - \mathrm{M}}{\mathrm{m} + \mathrm{M}}\right)}^2$u²

∴     $\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{i}}} = {\left(\frac{\mathrm{m} - \mathrm{M}}{\mathrm{m} + \mathrm{M}}\right)}^2$

The fraction of kinetic energy transferred from the neutron to the carbon atom is

      f = $\frac{4\mathrm{m} \left(12 \mathrm{m}\right)}{{\left( \mathrm{m} + 12\mathrm{m} \right)}^2}$

      f = $\frac{48}{169}$

C.

If assertion is true but reason is false.

When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the coulomb repulsive behaviour, it is called thermonuclear fusion.

It is clean source of energy but energy released in one fusion is much less than a single uranium fission.

D.

18 and 8

          ₇X¹⁵  + ₂He⁴  →   ₁H¹  +  _ZY^A

                     Alpha        Proton
                     particle

According to the law of conservation of charge, we get

          7 + 2 = 1 + Z

⇒                Z = 8

According to the law of conservation of mass, we get

          15 + 4 = 1 + A

⇒             A = 18

Hence, the resulting atom has mass number 18  and  atomic number 8.

A.

40 mA

Current increases with increase in voltage.

           Power = voltage × current

When voltage is maximum then current will be maximum.

Therefore maximum ower dissipation occurs at maximum voltage

           P_max = V_max I_Zmax

            P_max = 364 mW

                     = 364  × 10^-3 W

          V_max = 9.1 V

The maximum permissible current is

            I_Zmax = $\frac{{\mathrm{P}}_{\max }}{{\mathrm{V}}_{\max }}$

                     = $\frac{364 \times {10}^{-3}}{9.1}$

               I_Zmax = 40 mA

B.

If both assertion and reason are true but reason is not the correct explanation of assertion.

In heavy nuclei, the repulsive forces between the protons are more effective the nuclear attractive forces and these forces reduce the stability of the nucleus.

A.

$\frac{1}{16}$

Half-life T = 1600 years

Time of decay t = 6400 years

Number of half-lives

             n = $\frac{\mathrm{t}}{\mathrm{T}}$

                = $\frac{6400}{1600}$

            n = 4

Fraction undeecayed = $\frac{\mathrm{N}}{{\mathrm{N}}_0}$

                                = ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

                                = ${\left(\frac{1}{2}\right)}^4$

 Fraction undeecayed = $\frac{1}{16}$

B.

increasing the number of turns in the coils

Mutual inductance of coils

           M = $\frac{{\mathrm{\mu }}_0 {\mathrm{\mu }}_{\mathrm{r}} {\mathrm{N}}_1 {\mathrm{N}}_2 \mathrm{A}}{\mathrm{I}}$

It is clear that mutual inductance of coils can be increased by increasing the number of turns in the coils.

B.

If both assertion and reason are true but reason is not the correct explanation of assertion

The nuclear binding energy is termed as energy balance in processes in which the nucleus splits into fragments composed of more than one nucleon. If new binding energy is available when light nuclei fuse ( nuclear fusion), or when heavy nuclei split ( nuclear fission), either process can result in the release of binding energy.