NEET Practice Questions (MCQs) with Answers & Solutions

(d) As the transition n = 4 and n = 3 , results in UV radiation and infrared radiation involves smaller amounts of energy UV. So we require a transition involving initial values of n greater than 4 e.g. 5$\to$4.

(a) Potential energy $\mathrm{U}=\mathrm{eV}={\mathrm{eV}}_0\ln \frac{\mathrm{r}}{{\mathrm{r}}_0}$ 
$\therefore$ Force $\mathrm{F}=-\left|\frac{\mathrm{dU}}{\mathrm{dr}}\right|=\frac{{\mathrm{eV}}_0}{\mathrm{r}}$ .
$\therefore$ The force will provide the necessary centripetal force. Hence $\frac{{\mathrm{mv}}^2}{\mathrm{r}}=\frac{{\mathrm{eV}}_0}{\mathrm{r}}\Rightarrow \mathrm{v}=\sqrt{\frac{{\mathrm{eV}}_0}{\mathrm{m}}}$ …..(i)
and $\mathrm{mvr}=\frac{\mathrm{nh}}{2\mathrm{\pi }}$               …..(ii)
From equation (i) and(ii) $\mathrm{mr}=\left(\frac{\mathrm{nh}}{2\mathrm{\pi }}\right)\sqrt{\frac{\mathrm{m}}{{\mathrm{eV}}_0}}$  or r ∝ n

(d) ${\mathrm{r}}_{\mathrm{m}}=\frac{{\mathrm{m}}^2}{\mathrm{Z}}\left(0.53 \stackrel{0}{\mathrm{A}}\right)=\left(\mathrm{n}\times 0.53 \stackrel{0}{\mathrm{A}}\right)\Rightarrow \frac{{\mathrm{m}}^2}{\mathrm{Z}}=\mathrm{n}$

m = 5 for ${}_{100}{}^{257}\mathrm{Fm}$ (the outermost shell)

and z = 100$\Rightarrow \mathrm{n}=\frac{5^2}{100}=\frac{1}{4}$

$$\begin{gathered} By formula N=N_0{\left(\frac{1}{2}\right)}^{t/T} \\ or {10}^4=8\times {10}^4{\left(\frac{1}{2}\right)}^{t/3} \\ or \left(\frac{1}{8}\right)={\left(\frac{1}{2}\right)}^{t/3} \\ or {\left(\frac{1}{2}\right)}^3={\left(\frac{1}{2}\right)}^{t/3}\Rightarrow 3=\frac{t}{3} \\ Hence, t=9 yr \end{gathered}$$

Suppose x $\alpha$-particles and y $\beta$-particles are emitted

So, change in mass no. is given by

4x = 200 - 168 = 32

x = 8

and change in atomic no. is given by

2x - y = 90 - 80 = 10

Putting value of x 

or 2$\times$8 - y = 10 

So, no. of $\beta$-particles y = 6

no. of $\alpha$-paritcles x = 8

Alternative

$$\begin{gathered} {}_{200}X_{90}\to {}_{168}Y_{80} \\ As, {}_{200}X_{90}\to \left(n_{2}H{e}^4\right)+m\left({}_{-1}\beta _0\right)+{}_{168}Y_{80} \\ therefore, in this reaction \\ 200=4n+168 or n=\frac{200-168}{4}=8 \\ Also, 90=2n-m+80 \\ or m=2n+80-90 \\ =2\times 8+80-90=6 \\ Thus, respective number of \alpha and \beta -particles will be 8 and 6 \end{gathered}$$

$\frac{7}{8}th$ 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.

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