Physics MCQs for NEET — Practice Questions with Answers

(a) Magnetic moment of an electron : $\mu =\frac{neh}{4{\mathrm{\pi m}}_{\mathrm{e}}}$

(d) Bohr's model assumptions-

(i) The electrons in a hydrogen atom travels around the nucleus in a circular orbit

(ii) The nucleus is of infinite mass and is at rest

(iii) Electrons in a quantized orbit will not radiate energy

(iv) Mass of electron remains constant

(b) Protons and neutrons are the constituent particles of nucleus.

(a) Using $∆\mathrm{E}\propto {\mathrm{Z}}^2$ a                  ($\therefore$ ${\mathrm{n}}_1$ and ${\mathrm{n}}_2$ are same)

$\Rightarrow \frac{\mathrm{hc}}{\mathrm{\lambda }}\propto {\mathrm{Z}}^2\Rightarrow {\mathrm{\lambda Z}}^2= \mathrm{constant}$

$\Rightarrow {\mathrm{\lambda }}_1{\mathrm{Z}}_1^2={\mathrm{\lambda }}_2{\mathrm{Z}}_2^2={\mathrm{\lambda }}_3{\mathrm{Z}}_3^2={\mathrm{\lambda }}_4{\mathrm{Z}}_4^2$

$$\begin{gathered} \Rightarrow {\mathrm{\lambda }}_1\times 1={\mathrm{\lambda }}_2\times 1^2={\mathrm{\lambda }}_3\times 2^2={\mathrm{\lambda }}_4\times 3^3 \\ \Rightarrow {\mathrm{\lambda }}_1={\mathrm{\lambda }}_2=4{\mathrm{\lambda }}_3=9{\mathrm{\lambda }}_4 \end{gathered}$$

(c) In hydrogen atom ${\mathrm{E}}_{\mathrm{n}}=-\frac{\mathrm{Rhc}}{{\mathrm{n}}^2}$
Also ${\mathrm{E}}_{\mathrm{n}}\propto \mathrm{m}$ ; where m is the mass of the electron. Here the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in nth orbit will be given by ${\mathrm{E}}_{\mathrm{n}}=-\frac{2\mathrm{Rhc}}{{\mathrm{n}}^2}$
The longest wavelength ${\mathrm{\lambda }}_{\max }$(or minimum energy) photon will correspond to the transition of particle from n = 3 to n = 2 $\Rightarrow \frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\max }}={\mathrm{E}}_3-{\mathrm{E}}_2=2\mathrm{Rhc}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$
This gives ${\mathrm{\lambda }}_{\max }=\frac{18}{5\mathrm{R}}$.

(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

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