Physics MCQs for NEET — Practice Questions with Answers

The time required for the number of parent nuclei to fall to 50% is called half-life T and may be related to disintegration constant $\lambda$ as follows.

Since, $$\begin{gathered} 0.5 N_0=N_0{e}^{-\lambda T} \\ [N=Final no. of nuclei=0.5 N_0] \\ [\lambda =decay cons\tan t] \\ {N}_0=Initial no. of nuclei \\ \lambda =decay cons\tan t \\ We have, \lambda T={\log }_{e}2 \\ \therefore \lambda =\frac{{\log }_e 2}{T} \end{gathered}$$

Alpha particle is a positive particle. An alpha particle has $3.2\times {10}^{-19} C$ charge twice the negative charge of an electron. The mass of an $\alpha -particle$ is $6.645\times {10}^{-27}$ kg  which is equal to mass of helium nucleus. When two electrons are emitted by a helium atom, a nucleus of helium remains which has charge equal to that of two electrons. Actuatlly alpha ($\alpha$) particle is a nucleus of helium. Hence, it is also called as doubly-ionised helium atom.

The activity of a radioactive substance is defined as the rate at which the nuclei of its atoms in the sample disintegrate. In two half-lives, the activity becomes one-fourth. Two months is 2 half-life period. The activity, two months earlier was $2\times 2^2$= 8 microcurie.

Note:- The activity of a radioactive sample is called one curie, if it undergoes $3.7\times {10}^{10}$ disintegrations per second.

Nuclear forces act between a pair of neutrons, a pair of protons and also between a neutron-proton pair, with the same strength. This shows that nuclear forces are independent of charge.

Number of atoms left after n half-lives is given by

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

Number of half-lives,

$$\begin{gathered} n=\frac{t}{T}=\frac{6400}{800}=8 \\ \therefore \frac{N}{N_0}={\left(\frac{1}{2}\right)}^8=\frac{1}{256} \end{gathered}$$

Alternative

Let the initial part be unity

so, after 800 years, it will remain = $\frac{1}{2}$

after 1600 years, it will remain = $\frac{1}{4}$

after 2400 years, it will remain = $\frac{1}{8}$

after 3200 years, it will remain = $\frac{1}{16}$

after 4000 years, it will remain = $\frac{1}{32}$

after 5600 years, it will remain = $\frac{1}{64}$

after 5600 years, it will remain = $\frac{1}{128}$

after 6400 years, it will remain = $\frac{1}{256}$

$$\begin{gathered} R=R_0{\left(A\right)}^{1/3} \\ {R}_{Al}=R_0{\left(27\right)}^{1/3}=3R_0 \\ {R}_{Te}=R_0{\left(125\right)}^{1/3}=5R_0=\frac{5}{3}{R}_{Al} \end{gathered}$$

When a nucleus emits an alpha particle, its mass number decreases by 4 and charge/atomic no. decreases by 2. In $\beta$-particle emission, mass remains same but atomic number is increased by one. In $\gamma$-decay, daughter nucleus has the same charge number and same mass number as those of parent nucleus. Hence, sequence is 

${}_{Z}X_A\stackrel{\beta }{\to }{}_{Z+1}Y_A\stackrel{\alpha }{\to }{}_{Z-1}K_{A-4}\stackrel{\gamma }{\to }{}_{Z-1}K_{A-4}$

Mass of proton = mass of antiproton

$=1.67\times {10}^{-27} kg=1 amu$

Energy equivalent to 1 amu = 931 MeV

So energy equivalent to 2 amu = $2\times 931 MeV$

$$\begin{gathered} =1862\times {10}^6\times 1.6\times {10}^{-19} \\ =2.97\times {10}^{-10} J=3\times {10}^{-10} J \end{gathered}$$

The relation between nuclei radius (R) and mass number (A) is given  by

$$\begin{gathered} R \alpha A^{1/3}........\left(i\right) \\ or A \alpha R^3 \\ or \frac{A_1}{A_2}={\left(\frac{R_1}{R_2}\right)}^3 \\ Given, R_1=R,R_2=\frac{R}{2}, A=56 \\ \therefore \frac{56}{A_2}={\left(\frac{R}{R/2}\right)}^3=2^3=8 \\ or A_2=\frac{56}{8}=7 \\ Thus, required stable nucleus will be L{i}^7 \end{gathered}$$

Density of nuclear matter is the ratio of mass of nucleus and its volume. If m is average mass of a nucleon and R is the nuclear radius, then mass of nucleus = mA, where A is the mass number of the element.

$Volume of nucleus=\frac{4}{3}\pi R^3$

_{$And R=R_0{\left(A\right)}^{1/3}$}

_{$\Rightarrow V=\frac{4}{3}\pi {\left(R_0\right)}^{3}A$}

As density of nuclear matter = $\frac{mass of nucleus}{volume of nucleus}$

$$\begin{gathered} \rho =\frac{mA}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}_0^3\mathrm{A}} \\ \therefore \rho =\frac{3m}{4\pi R_0^3} \end{gathered}$$

As m and $R_0$ are constants, therefore density $\rho$ of nuclear matter is constant.