NEET Practice Questions (MCQs) with Answers & Solutions

Conceptual

α-particles are charged particles, so they can deflect by electric field.

(b) Let initial number of nuclei in A and B is $N_O$

Number of nuclei of A after time t is 

               $N_A=N_{0e^{-8\lambda t}}$                          ...(i)

Similarly, number of nuclei of A after time t is   

             $N_B=N_0{e}^{-\lambda t}$                          ...(ii)

It is given that 

$\frac{N_A}{N_B}=\frac{1}{e} \left|\therefore N_B>N_A\right|$

Now,from Eqs.(i)  and (ii)

$\frac{e^{-B\lambda t}}{e^{-\lambda t}}=\frac{1}{e}$

Rearranging 

$\Rightarrow e^{-1}=e^{-7\lambda t} \Rightarrow 7\lambda t=1$

$\Rightarrow Time \left(t\right)=\frac{1}{7\lambda }$

(d) Key idea Half-life of a radioactive substance is $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$$\propto$log $\left(\frac{N_o}{N}\right)$

Given,     $$\begin{gathered} N_1=0.6N_{0 } \left(\because 40\% decay\right) \\ {N}_2=0.15N_{0 } \left(\because 85\% decay\right) \end{gathered}$$

Putting these in the formula

        $\frac{N_2}{N_1}=\frac{0.15N_0}{0.6N_0}=\frac{1}{4}={\left(\frac{1}{2}\right)}^2$

So, two half-life periods has passed.

Thus, time taken=2$\times t_{1/2}=2\times 30=60$ min

(b)

 $$\begin{gathered} Radius of nucleus is given by- \\ R=R_0{A}^{1/3} \\ \Rightarrow R \alpha A^{1/3} \\ \frac{R_{Al}}{R_{Te}}={\left(\frac{27}{125}\right)}^{1/3}=\frac{3}{5} \\ \Rightarrow R_{Te}=\frac{5}{3}{R}_{Al} \end{gathered}$$

U^{238
   92}
->Th²³⁸ + He⁴
        92        2
According to law of conservation of linear momentum, we have.

|P_Th|=|P_He|=P

=>As, kinetic energy of an element,

KE=P2/2m

where,m is mass of an element 

Thus,KE∝1/M

So, M_HE<M_Th=>K_He>K_Th

The binding energy for ${}_1{}^1\mathrm{H}$ is around zero and also not given in the question so we can ignore it.

Q=2(4x7.06)-7x(5.60)

=(8x7.06)-(7x5.60)

=(56.48-39.2)MeV

=17.28MeV≈17.3MeV

Ratio of X:Y is given=1:7 

m_x/m_y=1/7

=> 7m_x/m_y

=> Let the initial total mass is m.

=> m_x+m_y=m => m_y/7+m_y=m

=> 8m_y/7=m

=> m_y=7/8m

only 1/8 part remains 

=> 1->1/2->1/4->1/8
           T/2    T/2   T/2

So,time taken to become 1/8 unstable part

=3 x T_1/2 = 3 x 1.4 x 10⁹ =4.2 x 10⁹ y

(b) As N/N_o=(1/2)ⁿ 

N/N_o=(1/2)³=1/8

Number of half lices=3

=> T=20yr  

∴ T=t/n or t=T x n

=20 x 3yr=60yr

(c) Here,Δm=0.02866U

∴ Energy liberated per nucleon

=0.02866 x 931/4

=26.7/4 MeV

=6.675MeV