NEET Practice Questions (MCQs) with Answers & Solutions

$$\begin{gathered} A=\left|\frac{-dN}{dt}\right|=counting rate=activity=N\lambda \\ {A}_0=1600 at t=0 A=100 at t=8s \\ \frac{A}{A_0}=\frac{100}{1600}=\frac{1}{16}={\left(\frac{1}{2}\right)}^{44}=\frac{N\lambda }{N_0\lambda }=\frac{N}{N_0} \\ \therefore \frac{N}{N_0}={\left(\frac{1}{2}\right)}^4={\left(\frac{1}{2}\right)}^{t/T} \\ \therefore \frac{t}{T}=4, t=8 s; T=half life=2 s; t\text{'}=6 \\ second=3 half life \\ \therefore \frac{A\text{'}}{A_0}=\frac{N\text{'}\lambda }{N_0\lambda }=\frac{N\text{'}}{N_0}={\left(\frac{1}{2}\right)}^3=\frac{1}{8} \\ \frac{A\text{'}}{1600}=\frac{1}{8}; A\text{'}=\frac{1600}{8}=200 \end{gathered}$$

$N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}=N_0{\left(\frac{1}{2}\right)}^{\frac{15}{5}}=\frac{N_0}{8}$

In 2 s only 90% nuclei are left behind. Thus, in the next 2 s, 90% of 900 or 810 nuclei will be left.

In the case of formation of a nucleus, the evolution of energy equal to the binding energy of the nucleus takes place due to disappearance of a fraction of the total mass. If the quantity of mass disappearing is $∆$m, then the binding energy is BE = $∆$m$c^2$

From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the consituent neutrons and protons. We can then write.

$∆m=Z{m}_p+N{m}_n-m(A, Z)$

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

BE = [Z$m_p$+ N$m_n$- m(A, Z)$c^2$

BE = [Z$m_p$+ (A-Z)$m_n$- m(A, Z)$c^2$

where, N = A -Z = number of neutrons

The equation ${}_{Z}X_A\to {}_{Z+1}Y_A + {}_{-1}e_0+\overline{\nu }$ represents  $\beta$-emission (since  $\beta$-ray electron is emitted)  

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 remained undecayed

$N=N_0{\left(\frac{1}{2}\right)}^n$

where $N_0$ is initial number of nuclei and N is number of nuclei

So for two different cases (A) and (B),

$$\begin{gathered} \frac{N_A}{N_B}=\frac{{\left({\displaystyle \frac{1}{2}}\right)}^{n_A}}{{\left({\displaystyle \frac{1}{2}}\right)}^{n_B}} or \frac{N_A}{N_B}=\frac{{\left({\displaystyle \frac{1}{2}}\right)}^4}{{\left({\displaystyle \frac{1}{2}}\right)}^2}=\frac{\left({\displaystyle \frac{1}{16}}\right)}{\left({\displaystyle \frac{1}{4}}\right)} \\ or \frac{N_A}{N_B}=\frac{1}{4} \end{gathered}$$

The radioactive isotope ${}_{27}{}^{60}Co$ decays into the stable isotope ${}_{28}{}^{60}Ni$ by emitting two gamma rays with energies 1.17 MeV and 1.33 MeV in succession.

Order of Radius of atom $\approx {10}^{-10} m$

Order of Radius of nucleus $\approx {10}^{-15} m$

Ratio of volume of atom to volume of nucleus

$$\begin{gathered} =\frac{Volume of atom}{Volume of nucleus} \\ =\frac{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}_1^3}{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}_2^3}={\left(\frac{{10}^{-10}}{{10}^{-15}}\right)}^3={10}^{15} \end{gathered}$$

${}_{72}A_{180}\stackrel{\alpha }{\to }{}_{70}A_1^{176}\stackrel{\beta }{\to }{}_{71}A_2^{176}\stackrel{\alpha }{\to }{}_{69}A_3^{172}\stackrel{\gamma }{\to }{}_{69}A_4^{172}$

$$\begin{gathered} Fraction undecayed after two half-lives ={\left(\frac{1}{2}\right)}^2=\frac{1}{4} \\ Fraction decayed after two half-lives=\frac{3}{4} \end{gathered}$$