Dual Nature of Matter and Radiation MCQ Questions & Answers in Modern Physics | Physics

$$\eqalign{ & \lambda = \frac{h}{{\sqrt {2mqV} }}\,\,{\text{and}}\,\,mV = {\text{constant}} \cr & 1837\,V' = 1 \times V\,\,{\text{or}}\,\,V' = \frac{V}{{1837}}volt \cr} $$

$$\eqalign{ & {\text{Power}} \propto {\text{No}}{\text{. of}}\,{\text{electrons emitted}}\,\left( N \right) \cr & P \propto \frac{1}{{{r^2}}} \Rightarrow N \propto \frac{1}{{{r^2}}} \cr} $$

de-Broglie wavelength is given by
$$\lambda = \frac{h}{{mv}}$$
For same velocity, $$\lambda \propto \frac{1}{m}$$
Out of the given particles, the mass of $$\beta $$-particle which is a fast moving electron, is minimum. Thus, de-Broglie wavelength is maximum for $$\beta $$-particle.

According to Einstein's photoelectric equation,
$$E = {K_{\max }} + \phi $$
where, $${K_{\max }}$$  is maximum kinetic energy of emitted electron and $$\phi $$ is work function of an electron.
$$\eqalign{ & {K_{\max }} = E - \phi = hv - \phi \cr & {K_{\max }} = \frac{{hc}}{\lambda } - \phi \,.......\left( {\text{i}} \right) \cr} $$
Similarly, in second case, maximum kinetic energy of emitted electron is 3 times that in first case, we get
$$3{K_{\max }} = \frac{{hc}}{{\frac{\lambda }{2}}} - \phi \,.......\left( {{\text{ii}}} \right)$$
Solving Eqs. (i) and (ii), we get work function of an emitted electron from a metal surface.
$$\phi = \frac{{hc}}{{2\lambda }}$$

According to De-broglie $$p = \frac{h}{\lambda }{\text{ or}}\,\,P = \frac{1}{\lambda }$$
$$P \propto \frac{1}{\lambda }$$  represents rectangular hyperbola.

For photon,
$${\lambda _2} = \frac{{hc}}{E}\,......\left( {\text{i}} \right)$$
For proton, $$p = \sqrt {2mE} $$
$$\eqalign{ & {\lambda _1} = \frac{h}{p} = \frac{h}{{\sqrt {2mE} }}\,......\left( {{\text{ii}}} \right) \cr & \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{hc}}{{E \times \frac{h}{{\sqrt {2mE} }}}} \propto {E^{ - \frac{1}{2}}} \cr} $$

$$\eqalign{ & \Delta E.\Delta t \sim h \cr & E = \frac{{hc}}{\lambda } \cr & \Delta E = \frac{{hc}}{\lambda }\frac{{\Delta \lambda }}{\lambda } \cr & \therefore \frac{{hc}}{\lambda }\frac{{\Delta \lambda }}{\lambda }.\Delta t \sim h \cr & {\text{Now,}}\,c = v\lambda \cr & v\Delta \lambda + \lambda \Delta v = 0 \cr & vD\lambda = - \lambda \Delta v \cr & \therefore \frac{{\Delta \lambda }}{\lambda } = - \frac{{\Delta v}}{v} \cr & \therefore \frac{{ch}}{\lambda }\frac{{\Delta v}}{v}\Delta t \sim h \cr & \therefore \frac{{\Delta v}}{v} \sim \frac{h}{{\Delta t}}.\frac{\lambda }{{hc}} \sim \frac{\lambda }{{\Delta tc}} = \frac{{600 \times {{10}^{ - 9}}}}{{8 \times {{10}^{ - 9}} \times 3 \times {{10}^8}}} \cr & \frac{{\Delta v}}{v} \sim {10^{ - 6}} \cr} $$

$$\eqalign{ & V = \frac{{12400}}{{{\lambda _{\min }}\left( {\mathop {\text{A}}\limits^ \circ } \right)}}Volt \cr & \Rightarrow V = \frac{{12400}}{2} = 6.2\,KV \cr} $$

$$\eqalign{ & \frac{{{E_1}}}{{{E_2}}} = \frac{{hc}}{{{\lambda _1}}}\frac{{{\lambda _2}}}{{hc}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} \cr & \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{50}}{1} \cr} $$

At low pressures, the density of the gas decreases, the mean free path of the gas molecules becomes large. Now, under the effect of external high voltage, the ions acquire sufficient energy before they collide with molecules, causing further ionisation. Due to it, the number of ions in the gas increases and it becomes a conductor.