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

Stopping potential is the negative potential applied to stop the electrons having maximum kinetic energy. Therefore, stopping potential will be 4 volt.

According to Einstein photoelectric equation
$$\eqalign{ & KE = E = {W_0}\,\,\left[ {_{{W_0} = \,{\text{work}}\,{\text{function}}}^{E\, = \,\,{\text{energy incidented}}}} \right] \cr & {\text{Here,}}\,\,E = 6.2\,eV \cr & {W_0} = 4.2\,eV \cr & \therefore KE = 6.2 - 4.2 = 2.0\,eV \cr & = 2 \times 1.6 \times {10^{ - 19}} \cr & = 3.2 \times {10^{ - 19}}\,J \cr} $$

Work function of a metal is $${W_0} = h{\nu _0}$$
$${\text{or}}\,{W_0} = \frac{{hc}}{{{\lambda _0}}}$$
Considering the two situation work function is given by
$$\therefore \frac{{{{\left( {{W_0}} \right)}_1}}}{{{{\left( {{W_0}} \right)}_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{600}}{{300}} = 2$$

The energy of a photon of a radiation of frequency $$\nu $$ and wavelength $$\lambda $$ is
$$E = h\nu = \frac{{hc}}{\lambda }\,.......\left( {\text{i}} \right)$$
If photon is considered to be a particle of mass $$m,$$ the energy associated with it, according to Einstein mass energy relation, is given by
$$E = m{c^2}\,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii),
$$\eqalign{ & h\nu = m{c^2} \cr & {\text{or}}\,\,m = \frac{{h\nu }}{{{c^2}}} \cr & \Rightarrow \frac{{h\nu }}{c} = mc \cr & {\text{As}}\,\,mc = p\,\,\left( {p = {\text{momentum of photon}}} \right) \cr & {\text{So,}}\,\,\nu = \frac{{cp}}{h} = \frac{{3 \times {{10}^8} \times 3.3 \times {{10}^{ - 29}}}}{{6.6 \times {{10}^{ - 34}}}} \cr & = 1.5 \times {10^{13}}Hz \cr} $$

In thermionic emission and other types of emission, ions emitted are atoms that has lost or gained electrons having negative or positive charge respectively. Thus, the nature of ions knocked out from hot surfaces, are electrons.

Maximum $$KE$$  depends on the frequency of incident radiation, not on intensity.

By Einstein equation
$$\eqalign{ & hf = {W_0} + e{V_0}. \cr & {W_0} = \frac{{hc}}{\lambda } - e{V_0} \cr & = \frac{{\left( {6.63 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{4950 \times {{10}^{ - 10}}}} - \left( {1.6 \times {{10}^{ - 19}}} \right) \times 0.6 \cr & = 3.12 \times {10^{ - 19}}J \cr} $$
For the second surface
$$\eqalign{ & hf' = {W_0} + e{V_0}^\prime \cr & {\text{or}}\,\,\frac{{hc}}{{\lambda '}} = 3.12 \times {10^{ - 19}} + 1.6 \times {10^{ - 19}} \times 1.1 \cr & \therefore \lambda ' = 4077\,\mathop {\text{A}}\limits^ \circ \cr} $$

In 1905, Einstein realized that the photoelectric effect could be understood if the energy in light is not spread out over wavefronts but is concentrated in small packets, or photons. Each photon of light of frequency $$\nu $$ has the energy $$h\nu .$$ Thus, Einstein's work on photoelectric effect gives support to $$E = h\nu .$$

Intensity of light is inversely proportional to square of distance of source
$${\text{i}}{\text{.e}}{\text{.}}\,I \propto \frac{1}{{{d^2}}}$$
For two different situations, $$\frac{{{I_2}}}{{{I_1}}} = \frac{{{{\left( {{d_1}} \right)}^2}}}{{{{\left( {{d_2}} \right)}^2}}}$$
Given, $${d_1} = 0.5\,m,{d_2} = 1.0\,m$$
Therefore, $$\frac{{{I_2}}}{{{I_1}}} = \frac{{{{\left( {0.5} \right)}^2}}}{{{{\left( 1 \right)}^2}}} = \frac{1}{4}$$
Since, number of photoelectrons emitted per second is directly proportional to intensity, so number of electrons emitted would decrease by factor of 4.

Energy of photon is $$E = h\nu = \frac{{hc}}{\lambda }$$
As, $$\frac{{hc}}{\lambda } = {10^3}eV$$
$$\therefore h = \frac{{{{10}^3}\lambda }}{c}\,......\left( {\text{i}} \right)$$
And for 2^nd case as given in question
$$h\nu = {10^6}eV\,.......\left( {{\text{ii}}} \right)$$
Putting value of $$h$$ in Eq. (ii),
$$\eqalign{ & \frac{{{{10}^3}\lambda }}{c}\nu = {10^6} \cr & \therefore \nu = \frac{{{{10}^3}c}}{\lambda } \cr & = \frac{{{{10}^3} \times 3 \times {{10}^8}}}{{1.24 \times {{10}^{ - 9}}}}\,\,\left[ {\because \lambda = 1.24 \times {{10}^{ - 9}}m} \right] \cr & = 2.4 \times {10^{20}}Hz \cr} $$