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

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

In a discharge tube, after being accelerated though a high potential difference the ions in the gas strike the cathode with huge kinetic energy. This collision liberates electrons from the cathode. These free electrons can further liberate ions from gas molecules through collisions. The positive ions are attracted towards the cathode and negatively charged electrons move towards anode. Thus, ionisation of gas results.

Wavelength of electron is given by
$${\lambda _e} = \frac{h}{{{p_e}}} = \frac{h}{{\sqrt {2mE} }}\,\,\left[ {\because {p_e} = \sqrt {2mE} } \right]$$
and for photon,
$$\eqalign{ & {\lambda _p} = \frac{{hc}}{E} \Rightarrow \lambda _e^2 = \frac{{{h^2}}}{{2mE}}\,\,{\text{or}}\,\,E = \frac{{hc}}{{{\lambda _p}}} \cr & \therefore \lambda _e^2 = \frac{{{h^2}}}{{2m \cdot \frac{{hc}}{{{\lambda _p}}}}} \Rightarrow \lambda _e^2 = \frac{{{h^2}}}{{2mhc}}{\lambda _p} \cr & \Rightarrow \lambda _e^2 \propto {\lambda _p} \cr} $$

$$\eqalign{ & {\text{Use}}\,\,n = \frac{I}{{hv}} = \frac{{I\lambda }}{{hc}} \cr & \therefore \frac{{{n_{UV}}}}{{{n_{IR}}}} = \frac{{{\lambda _{UV}}}}{{{\lambda _{IR}}}} = \frac{{400}}{{700}} = \frac{4}{7} = 0.57 \cr} $$

$$\eqalign{ & {P_{in}} = 25\,W,\lambda = 6600\,\mathop {\text{A}}\limits^ \circ = 600 \times {10^{ - 10}}m \cr & nhv = P \cr} $$
⇒ Number of photons emitted/sec,
$$\eqalign{ & n = \frac{P}{{\frac{{hc}}{\lambda }}} = \frac{{P\lambda }}{{hc}} = \frac{{25 \times 6600 \times {{10}^{ - 10}}}}{{6.64 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \cr & = 8.23 \times {10^{19}} = \frac{{25}}{3} \times {10^{19}} \cr} $$
$$3\% $$  of emitted photons are producing current
$$\therefore I = \frac{3}{{100}} \times ne = \frac{3}{{100}} \times \frac{{25}}{3} \times {10^{19}} \times 1.6 \times {10^{ - 19}} = 0.4\,A$$

Velocity at highest point $$ = u\sin \theta .$$
$$\therefore {\lambda _D} = \frac{h}{{mu\sin \theta }}\,\,\left( {{\text{Since }}\theta {\text{ is velocity wrt vertical}}} \right)$$

Momentum imparted per unit time $$= np$$
$$\eqalign{ & \Rightarrow F = \frac{{nh}}{\lambda } \cr & \therefore {\text{Acceleration}} = \frac{{nh}}{{m\lambda }} \cr} $$

No. of photons/sec $$ = \frac{{{\text{Energy incident on platinum surface per second}}}}{{{\text{Energy of one photon}}}}$$
No. of photons incident per second $$ = \frac{{2 \times 10 \times {{10}^{ - 4}}}}{{10.6 \times 1.6 \times {{10}^{ - 19}}}} = 1.18 \times {10^{14}}$$
As $$0.53\% $$  of incident photon can eject photoelectrons
$$\therefore $$ No. of photoelectrons ejected per second
$$ = 1.18 \times {10^{14}} \times \frac{{0.53}}{{100}} = 6.25 \times {10^{11}}$$
Minimum energy $$= 0\,eV,$$
Maximum energy $$ = \left( {10.6 - 5.6} \right)eV = 5\,eV$$

Einstein's photoelectric equation is given by
$$KE = E - {W_0}$$
As we know that
$$\eqalign{ & KE = \frac{1}{2}m{v^2}\,\,{\text{and}}\,\,E = \frac{{hc}}{\lambda } \cr & \therefore \frac{1}{2}m{v^2} = \frac{{hc}}{\lambda } - {W_0}......\left( {\text{i}} \right) \cr} $$
Suppose $${v'}$$ be the new speed, when $$\lambda $$ is changed to $$\frac{{3\lambda }}{4}.$$
The new equation can be written as
$$\frac{1}{2}m{{v'}^2} = \frac{{hc}}{{\left( {\frac{{3\lambda }}{4}} \right)}} - {W_0}$$
$${\text{or}}\,\,\frac{1}{2}m{{v'}^2} = \frac{4}{3}\frac{{hc}}{\lambda } - {W_0}\,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we get
$$\eqalign{ & \frac{{{{v'}^2}}}{{{v^2}}} = \frac{{\frac{4}{3}\frac{{hc}}{\lambda } - {W_0}}}{{\frac{{hc}}{\lambda } - {W_0}}} \cr & = \frac{{\frac{4}{3}\frac{{hc}}{\lambda } - \frac{4}{3}{W_0} + \frac{1}{3}{W_0}}}{{\frac{{hc}}{\lambda } - {W_0}}} \cr & = \frac{4}{3} + \frac{{{W_0}}}{{3\left( {\frac{{hc}}{\lambda } - {W_0}} \right)}} > \frac{4}{3} \cr & \therefore \frac{{v'}}{v} > \sqrt {\frac{4}{3}} \,\,{\text{or}}\,\,v' > \sqrt {\frac{4}{3}v} \cr} $$

If $$v$$ is the speed acquired by particle when accelerated under a potential difference $$V,$$ then
$$KE$$  gained by particle $$ = \frac{1}{2}m{v^2} = eV$$
or $$v = \sqrt {\frac{{2\,eV}}{m}} $$
So, for two different cases of the He-atom and H-atom,
$$\frac{{{v_{He}}}}{{{v_H}}} = \frac{{\sqrt {\frac{{2{{\left( e \right)}_{He}}V}}{{{m_{He}}}}} }}{{\sqrt {\frac{{2{{\left( e \right)}_H}V}}{{{m_H}}}} }}$$
As $${m_{He}} = 4\,{m_H}\,\,{\text{and}}\,\,{\left( e \right)_{He}} = 2{\left( e \right)_H}$$
$$\therefore \frac{{{v_{He}}}}{{{v_H}}} = \frac{1}{{\sqrt 2 }}$$