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

Energy of photon is given by
$$\eqalign{ & E = \frac{{hc}}{\lambda } = \frac{{12375}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}}\,eV\,\,\left[ {\because hc = 12375\,eV - \mathop {\text{A}}\limits^ \circ } \right] \cr & \therefore E = \frac{{12375}}{{5000}} \cr & = 2.48\,eV \cr} $$
According to Einstein’s photoelectric equation
$$\eqalign{ & KE = E - {W_0} = 2.48\,eV - 1.9\,eV \cr & = 0.58\,eV \cr} $$

Total current is due to electrons and positively charged ions
$$\eqalign{ & {\text{So,}}\,\,{\text{current}}\,\,{I_{net}} = {I_e} + {I_p} \cr & {\text{Also,}}\,\,{\text{current}}\,\,{I_{net}} = neA{v_d} \cr} $$
where, $${v_d}$$ is drift velocity, $$A$$ is area of cross-section, $$n$$ the number density of ions.
$$\eqalign{ & {\text{Given,}}\,\,n = 5 \times {10^7}/c{m^3} = 5 \times {10^{13}}/{m^3}, \cr & {v_e} = 0.4\,m/s \cr} $$
∴ Electron current $${I_e} = 5 \times {10^{13}} \times 1.6 \times {10^{ - 19}} \times A \times 0.4$$
Current due to positively charged ions
$$\eqalign{ & {I_p} = 5 \times {10^{13}} \times 1.6 \times {10^{ - 19}} \times A \times v\,\,\left[ {v = {\text{velocity of single}}\,{\text{charged}}\,{\text{ions}}} \right] \cr & {\text{As}}\,\,I = {I_e} + {I_p} = 5 \times {10^{13}} \times 1.6 \times {10^{ - 19}} \times A\left( {v + 0.4} \right) \cr & {\text{Given,}}\,\,4 \times {10^{ - 6}} \times A = 5 \times {10^{ - 6}} \times 1.6 \times A\left( {v + 0.4} \right) \cr & \Rightarrow 0.5 = v + 0.4 \cr & \therefore v = 0.1\,m/s \cr} $$

Since, stopping potential is independent of distance hence new stopping potential will remain unchanged i.e., new stopping potential $$ = {V_0}.$$

In I^st case, when a metallic surface is illuminated with radiation of wavelength $$\lambda ,$$ the stopping potential is $$V.$$
So, photoelectric equation can be written as
$$eV = \frac{{hc}}{\lambda } - \frac{{hc}}{{{\lambda _0}}}\,......\left( {\text{i}} \right)$$
In II^nd case, when the same surface is illuminated with radiation of wavelength $$2\lambda ,$$  the stopping potential is $$\frac{V}{4}.$$ So, photoelectric equation can be written as
$$\eqalign{ & \frac{{eV}}{4} = \frac{{hc}}{{2\lambda }} - \frac{{hc}}{{{\lambda _0}}} \cr & \Rightarrow eV = \frac{{4hc}}{{2\lambda }} - \frac{{4hc}}{{{\lambda _0}}}\,......\left( {{\text{ii}}} \right) \cr} $$
From Eqs. (i) and (ii), we get
$$\eqalign{ & \Rightarrow \frac{{hc}}{\lambda } - \frac{{hc}}{{{\lambda _0}}} = \frac{{4hc}}{{2\lambda }} - \frac{{4hc}}{{{\lambda _0}}} \cr & \Rightarrow \frac{1}{\lambda } - \frac{1}{{{\lambda _0}}} = \frac{2}{\lambda } - \frac{4}{{{\lambda _0}}} \cr & \Rightarrow {\lambda _0} = 3\lambda \cr} $$

$$I = \frac{q}{t} = \frac{{ne}}{t}$$
No. of electrons striking the target per second
$$ = \frac{I}{e} = 2 \times {10^{16}}$$

Intensity of light source is inversely proportional to the distance $$\left( d \right)$$
i.e. $$I \propto \frac{1}{{{d^2}}}$$
When distance is doubled, intensity becomes one-fourth.
As number of photoelectrons $$ \propto $$ intensity, so number of photoelectrons is quarter of the initial number.

$$\eqalign{ & {\text{Energy in joule}}\,\left( E \right) = {\text{charge}} \times {\text{potential diff}}{\text{. in}}\,{\text{volt}} \cr & {E_{{\text{electron}}}} = {q_e}V{\text{and}}\,{E_{{\text{proton}}}} = {q_p}\,4V \cr} $$
de-Broglie wavelength $$\lambda = \frac{h}{P} = \frac{h}{{\sqrt {2mE} }}$$
$$\eqalign{ & {\lambda _e} = \frac{h}{{\sqrt {2{m_e}eV} }}\,\,{\text{and}}\,\,{\lambda _P} = \frac{h}{{\sqrt {2{m_P}e4V} }}\,\,\left( {\because {q_e} = {q_p}} \right) \cr & \therefore \frac{{{\lambda _e}}}{{{\lambda _P}}} = \frac{{\frac{h}{{\sqrt {2{m_e}eV} }}}}{{\frac{h}{{\sqrt {2{m_P}e4V} }}}} = \sqrt {\frac{{2{m_P}e4V}}{{2{m_e}eV}}} = 2\sqrt {\frac{{{m_P}}}{{{m_e}}}} \cr} $$

In the phenomenon of electric discharge through gases at low pressure, as the charged particles emitted from the cathode collides with the atoms of the gas, coloured glow appears in the tube.

$$\eqalign{ & {\text{As, energy of photon,}}\,\,E = h\nu \Rightarrow E = \frac{{hc}}{\lambda } \cr & \Rightarrow E = \frac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{500 \times {{10}^{ - 9}}}} \cr & \Rightarrow E = \frac{{0.0397 \times {{10}^{ - 34}} \times {{10}^8}}}{{{{10}^{ - 9}}}} = 0.0397 \times {10^{ - 21}}\,J \cr & = \frac{{0.0397 \times {{10}^{ - 21}}}}{{1.6 \times {{10}^{ - 19}}}} = 0.0248 \times {10^2}\,eV \cr & = 2.48\,eV \cr} $$
According to Einstein's photoelectric emission, we have
$$K{E_{\max }} = E - W = 2.48 - 2.28 = 0.2\,eV$$
For de-Broglie wavelength of the emitted electron,
$$\eqalign{ & {\lambda _{e\min }} = \frac{{12.27\,A}}{{\sqrt {K{E_{\max }}\left( {eV} \right)} }} = \frac{{12.27}}{{\sqrt {0.2} }} \cr & = 27.436\,\mathop {\text{A}}\limits^ \circ = 27.436 \times {10^{ - 10}}\;m \cr} $$
Thus, Minimum wavelength of the emitted electron is
$$\eqalign{ & {\lambda _{\min }} = 2.7436 \times {10^{ - 9}}\,m \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,\lambda \geqslant {\lambda _{\min }} \cr} $$

Number of emitted electrons
$$\eqalign{ & = \frac{{{{10}^{12}} \times 2 \times {{10}^{ - 4}} \times 25}}{{{{10}^5}}} = 5 \times {10^4}\,{\text{electrons}} \cr & q = + ne = 8 \times {10^{ - 15}}C \cr} $$