NEET Practice Questions (MCQs) with Answers & Solutions

According to law of radioactivity

$\frac{N}{N_0}=e^{-\lambda t}.....\left(i\right)$

$$\begin{gathered} \Rightarrow \frac{N_0}{N}=e^{\lambda t} \\ [N=final concentration] \\ [N_0=initial concentration] \\ [\lambda =decay cons\tan t] \end{gathered}$$

Taking logarithm on both sides of Eq. (i), we have

$$\begin{gathered} {\log }_e\left(\frac{N_0}{N}\right)={\log }_e\left(e^{\lambda t}\right) \\ =\lambda t {\log }_e e=\lambda t \end{gathered}$$

As we know that, ${\log }_e x=2.3026 {\log }_{10} x$

making substitution, we get 

$$\begin{gathered} \lambda =\frac{2.3026 {\log }_{10}\left({\displaystyle \frac{9750}{975}}\right)}{5} \\ [\because N_0=9750 counts/min and N=975 counts/min] \\ =\frac{2.3026}{5}{\log }_{10} 10 \\ =\frac{2.3026}{5} mi{n}^{-1} \\ =0.461 mi{n}^{-1} \end{gathered}$$

According to Einstein mass-energy equivalence is represented by E = m$c^2$

Taking mass, m = 1 amu = 1.66$\times {10}^{-27} kg$, and velocity of light in vaccum, c = 3.0$\times {10}^8 m/s$

We get $E=(1.66\times {10}^{-27})\times {(3\times 108)}^2 J$

$$\begin{gathered} =1.49\times {10}^{-10} J=\frac{1.49\times {10}^{-10}}{1.6\times {10}^{-13}}MeV \\ (\because 1 MeV=1.6\times 1.6\times {10}^{-13} J) \\ =931.25 MeV \end{gathered}$$

Hence, 1 amu$\approx$931 MeV

The source of solar energy is the nuclear fusion reaction occurring at the core of the Sun, where hydrogen nuclei fuse to form helium nuclei, releasing enormous amounts of energy in the process.

$$\begin{gathered} When the two samples are mixed, \\ N=N_1{e}^{-{\lambda }_{1}t}+N_2{e}^{-{\lambda }_{2}t} \\ Rate of decay=\frac{dN}{dt}=-{\lambda }_1{N}_1{e}^{-{\lambda }_{1}t}-{\lambda }_2{N}_2{e}^{-{\lambda }_{2}t} \end{gathered}$$

The reaction can be shown as

$$\begin{gathered} {}_{n}X_m\stackrel{\alpha \left({}_{2}H{e}^4\right)}{\to }{}_{n-2}Y_{m-4} \\ {}_{n-2}Y_{m-4}\stackrel{2\left({}_{-1}\beta _0\right)}{\to }{}_{n}X_{m-4} \end{gathered}$$

Thus, the resulting nucleus is the isotope of parent nucleus and is ${}_{n}X_{m-4}$.

The activity of a radioactive substance is

$$\begin{gathered} R=R_0{\left(\frac{1}{2}\right)}^n \\ [R=Final number] \\ [R_0=Initial number] \end{gathered}$$

Here, n = number of half-lives

$$\begin{gathered} =\frac{t}{T_{1/2}}=\frac{24}{6}=4 [T_{1/2}=Half life period] \\ and R=0.01 \mu Ci \\ So, 0.01 =R_0{\left(\frac{1}{2}\right)}^4 \\ or R_0=0.01\times {\left(2\right)}^4=0.01\times 16 \\ =0.16 \mu Ci \end{gathered}$$

$R=R_0{A}^{1/3}\Rightarrow R \alpha A^{1/3}$

Fraction undecayed is ${\left(\frac{1}{2}\right)}^5=\frac{1}{32}$

When a nucleus is formed, then the mass of nucleus is slightly less than the sum of masses of Z protons and N neutrons.

i.e., $m<\left(Z{m}_p+N{m}_n\right)$

Isotones are the nuclides which contain the same number of neutrons. In ${}_{6}C_{13} and {}_{7}N_{14}$, number of neutrons in carbon = 13 - 6 = 7 and number of neutrons in nitrogen = 14 - 7 = 7.

Note:-

No. of neutrons is given by

mass no.(A) - atomic no. (Z)