Dual Nature of Matter and Radiation MCQs for NEET — Physics Questions with Answers

Each moving particle consists of a wave. This wave is called matter-wave of De-Broglie wave.

$$\begin{gathered} h\nu =KE+\varphi \\ \varphi =\frac{hc}{\lambda }-KE=\frac{12400}{3100}-2.5=1.5eV \end{gathered}$$

$$\begin{gathered} \left(1\right) \\ \mathrm{Energy} \mathrm{of} \mathrm{photon} = \frac{\mathrm{hc}}{\mathrm{\lambda }} = \mathrm{p} \mathrm{x} \mathrm{c} \end{gathered}$$

$$\begin{gathered} \left(4\right) \\ \mathrm{\lambda } = \frac{\mathrm{h}}{\sqrt{2\mathrm{m}\left(\mathrm{K}.\mathrm{E}.\right)}} \Rightarrow \mathrm{\lambda } \propto \frac{1}{\sqrt{\mathrm{E}}} \end{gathered}$$

$$\begin{gathered} \left(3\right) \\ \mathrm{\lambda } = \sqrt{\frac{150}{\mathrm{V}}}\stackrel{o}{\mathrm{A}} \left(\mathrm{for} \mathrm{electron}\right) \\ \mathrm{\lambda } = \sqrt{\frac{150}{600}} = \frac{1}{2}\stackrel{o}{\mathrm{A}} \end{gathered}$$

$$\begin{gathered} \left(1\right) \\ \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{mv}} = \frac{6.6 \mathrm{x} {10}^{-34}}{1 \mathrm{x} 2 \mathrm{x} {10}^3} \\ = 3.3 \mathrm{x} {10}^{-37} = 3.3 \mathrm{x} {10}^{-27} \stackrel{\mathrm{o}}{\mathrm{A}} \end{gathered}$$

$$\begin{gathered} \left(3\right) \\ 0 = \overrightarrow{p_1} + \overrightarrow{p_2} \\ \left| \overrightarrow{p_2}\right| = \left| \overrightarrow{p_1}\right| \\ \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} \\ {\lambda }_1 = {\lambda }_2 \end{gathered}$$

(b) Thinking Processs de-Brogile wavelength associated with a moving particle can be given as 

 $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m(K.E)}}$

At thermal equilibrium,temperature of neutron and heavy water will be same. This common temperature is given as, T. Also, we know that, kinetic energy of a particle 

              $KE=\frac{p^2}{2m}$

    where, p=momentum of the particle 

 m=mass of the particle

Kinetic energy of the neutron is 

     $KE=\frac{3}{2}kT$

 $\therefore$ de-Brogile wavelength of the neutron 

      $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m(K.E)}}$

   $=\frac{h}{\sqrt{2m\times {\displaystyle \frac{3}{2}}kT}}=\frac{h}{\sqrt{3mkT}}$

(a) Key idea

Cut-off wavelength occurs when an incoming electron loses its complete energy in the collision. This energy appears in the form of X-rays.

Given, the mass of electrons=m

    de-Broglie wavelength=$\lambda$

So, kinetic energy, of electron = $\frac{p^2}{2m}$

                                        $=\frac{1}{2m}{\left(\frac{h}{\lambda }\right)}^2=\frac{h^2}{2m{\lambda }^2}$

Now, the maximum energy of a photon can be given by-

$$\begin{gathered} E=\frac{hc}{{\lambda }_0}=\frac{h^2}{2m{\lambda }^2} \\ \Rightarrow {\lambda }_0=\frac{hc\times 2{\lambda }^{2}m}{h^2} \\ = \frac{2mc{\lambda }^2}{h^{}} \end{gathered}$$

(c) In Ist case, when a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V.

So, photoelectric equation can be written as

              eV=$\frac{hc}{\lambda }-\frac{hc}{{\lambda }_o}$                        ...(i)

In IInd Case, when the same surface is illuminated with radition of wavelength 2$\lambda$, the stopping potential is $\frac{V}{4}$ So, photoelectric equation can be written as

                        $\frac{eV}{4}=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_0}$

$\Rightarrow eV=\frac{4hc}{2\lambda }-\frac{4hc}{{\lambda }_0}$                ...(ii)

From eqs, (i) and (ii) we get

$$\begin{gathered} \Rightarrow \frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=\frac{4hc}{2\lambda }-\frac{4hc}{{\lambda }_o} \\ \Rightarrow \frac{1}{\lambda }-\frac{1}{{\lambda }_o}= \frac{2}{\lambda }- \frac{4}{{\lambda }_o} \\ \Rightarrow {\lambda }_0=3\lambda \end{gathered}$$