Physics MCQs for NEET — Practice Questions with Answers

$\frac{7}{8}th$ decays means = $\frac{1}{8}th$ remains undecayed = $\frac{1}{8}={\left(\frac{1}{2}\right)}^3$

$\therefore 3 half life period=3\times 1622=4866 yrs$

${}_{2}H{e}^4$ contains 2 neutrons and 2 protons

So, mass of 2 protons = $2\times 1.0073=2.0146 u$

So, mass of 2 neutrons = $2\times 1.0087=2.0174 u$

Total mass of 2 protons and 2 neutrons = (2.0146+2.0174) u = 4.032 u

Mass of helium nucleus = 40015 u

Thus, mass defect is lacking of mass in forming the helium nucleus from 2 protons and 2 neutrons.

$∆$m = mass defect = (4.032-40015) u

Also we know that

1 u = 931 MeV

Hence, binding energy

$$\begin{gathered} ∆E=(∆m)\times 931 \\ =0.0305\times 931=28.4 MeV \end{gathered}$$

3 atoms of B $\to$One atom of A

$\therefore Q=E_a-3E_b$

Pauli suggested that after emission of $\beta$-particle (electron) a neutron is converted into a proton in a nucleus and in this reaction an electron and an antineutrino ($\nu$) will be formed. This reaction is represented as

$$\begin{gathered} {}_{0}n_1 \to {}_{1}H_1 + {}_{-1}\beta _0 +\overline{ \nu } \\ \left(neutron\right) \left(Proton\right) \left(Electron\right) \left(Antineutrino\right) \end{gathered}$$

Antineutrino is a particle whose mass is negligible and on which no charge is present.

Note:-

After emission of $\beta$-particle, the total number of particles (mass-number) in a nucleus remains uncharged but no. of neutrons reduces by 1 making the no. of protons (i.e. charge-number) to increase by 1.

$\frac{N}{N_0}=e^{-\lambda t}=e^{-\lambda \left(\frac{n}{\lambda }\right)}=\frac{1}{e^n}$

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 ${}_{n}X_{m-4}$.