Physics MCQs for NEET — Practice Questions with Answers

$$\begin{gathered} N=N_0{\left(\frac{1}{2}\right)}^2\Rightarrow \frac{N}{N_0}=\frac{1}{4} \\ Probability=1-\frac{N}{N_0}=1-\frac{1}{4}=\frac{3}{4} \end{gathered}$$

${}_{92}U_{235}+{}_{0}n_1\to {}_{38}S{r}^{90}+{}_{54}X{e}^{143}+3{}_{0}n_1$

If total atomic number on LHS = 92 + 0 = 92

Total atomic number on RHS = 38 + 54 + 0 = 92

Total mass number on LHS = 235 + 1 = 236

Total mass number on RHS = 90 + 143 + 3 $\times$ 1 = 236

Note :- For a nuclear reaction to be completed,the mass number and charge number on both sides should be same.

(a) Number of half-lives

        $n=\frac{t}{T_{1/2}}=\frac{18}{3}=6 \left[T_{1/2} = half life period\right]$

Amount remained after n half-lives

       $N=N_0{\left(\frac{1}{2}\right)}^n \left[\begin{array}{c}{N}_0=initial count\\ N=final count\end{array}\right]$

Given,   $N_0=300g$

$\therefore N=300{\left(\frac{1}{2}\right)}^6=300\times \frac{1}{64}=4.68 g$

$$\begin{gathered} Decay cons\tan t, \lambda =\frac{0.693}{T_{1/2}} \\ =\frac{0.693}{T_{1/2}}=\frac{0.693}{20}=0.03465 \\ Now, time of decay, \\ t=\frac{2.303}{\lambda } \log \frac{N_0}{N} \\ \Rightarrow t_1=\frac{2.303}{0.03465} \log \frac{100}{67}=11.6 min \\ and t_2=\frac{2.303}{0.03465} \log \frac{100}{33}=32 min \\ So, time difference between points of time \\ =t_1-t_2=32-11.6 \\ =20.4 min\approx 20 min \end{gathered}$$

${}_{1}H_2+{}_{1}H_2\to {}_{1}H_3+{}_{1}H_1$

The penetrating power of radiation is directly proportional to the energy of its photon.

Energy of photon = $\frac{hc}{\lambda }$ $\alpha \frac{1}{\lambda }$(wavelength)

$\therefore$ Penetrating power $\alpha \frac{1}{\lambda }$(wavelength)

Since, $\lambda$ is minimum for y-rays, so penetrating power is maximum for y-rays.

In fission process, when a parent nucleus breaks into daughter products, then some mass is lost in the form of energy. Thus, mass of fission products < mass of parent nucleus

$\Rightarrow \frac{Mass of fission products}{Mass of parent nucleus}< 1$

When a parent nucleus emits a $\beta$-particle mass number remains same because mass of electron is negligibly low. Atomic number is increased by one. The nucleus ${}_{48}C{d}^{115}$ after two successive $\beta$-decays will given ${}_{50}S{n}^{115}$.

In a nuclear process, energy is released if binding energy per nucleon of the daughter products gets increased. In nuclear fission reaction, total binding energy of products formed due to nuclear fission is more than the parent fissionable material.