NEET Practice Questions (MCQs) with Answers & Solutions

$\frac{3}{4}th$ active decay takes place in time

$t=2\left(T_{1/2}\right)\Rightarrow \frac{3}{4}=2\left(T_{1/2}\right)\Rightarrow T_{1/2}=\frac{3}{8} s$

Electronic configuration of iodine (53) is = 2, 8, 18, 18, 7

$\therefore$ Principal quantum number n = 5

Radius of nth orbit is given by 

$$\begin{gathered} r_n=r_0\left(\frac{n^2}{Z}\right) (Z=atomic number) \\ where r_0=0.53 \stackrel{0}{A}=0.53\times {10}^{-10} m \\ \therefore r_n=(0.53\times {10}^{-10})\times \frac{5^2}{53} \\ =2.5\times {10}^{-11} m \end{gathered}$$

Nuclear forces are the short range forces of attraction which hold together the nucleons (neutrons and protons) in the tiny nucleus of an atom, inspite of strong electrostatic forces of repulsion between protons. Nuclear forces are charge independent forces. They are the strongest forces in nature. The magnitude of nuclear forces is 100 times that of electrostatic forces and ${10}^{38}$ times that of gravitational forces between nucleons. They are operative upto distances of the order of a few fermi. They are non-central forces.

Experimental measurements show that volume of a nucleus is proportional to its mass number A. If R is the radius of the nucleus assumed to be spherical, then its volume and mass no. relation is given by

Volume (V) $\alpha$ mass no. (A)

$$\begin{gathered} or \left(\frac{4}{3}{\mathrm{\pi R}}^3\right) \alpha A \\ or R \alpha A^{1/3} \\ \therefore \frac{R_{Al}}{R_{Te}}=\frac{{\left(27\right)}^{1/3}}{{\left(125\right)}^{1/3}}=\frac{3}{5}=\frac{6}{10} \end{gathered}$$

Bragg's law for different of X-rays is

$2d \sin \theta =n\lambda$

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 = $\frac{1}{256}$