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

Energy of photon is given by
$$E = \frac{{hc}}{\lambda }\,\,.......\left( {\text{i}} \right)$$
where $$h$$ is Planck's constant, $$c$$ the velocity of light and $$\lambda $$ its wavelength.
de-Broglie wavelength is given by
$$\lambda = \frac{h}{p}\,......\left( {{\text{ii}}} \right)$$
$$p$$ being momentum of photon.
From Eqs. (i) and (ii), we have
$$E = \frac{{hc}}{{\frac{h}{p}}} = pc\,\,{\text{or}}\,\,p = \frac{E}{c}$$
Given, $$E = 1\,MeV = 1 \times {10^6} \times 1.6 \times {10^{ - 19}}J,$$
$$\eqalign{ & 1\,eV = 1.6 \times {10^{19}}J \cr & c = 3 \times {10^8}m/s \cr} $$
Hence, after putting numerical values, we obtain
$$\eqalign{ & p = \frac{{1 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{3 \times {{10}^8}}}kg - m/s \cr & = 5 \times {10^{ - 22}}\,kg - m/s \cr} $$

$$V$$ versus $$f$$ has a constant slope of $$\frac{h}{e},$$  so both lines must be parallel. Also, work function is equal to intercept on $$f$$-axis.

$$\eqalign{ & \frac{{{\lambda _{{m_2}}}}}{{{\lambda _{{m_1}}}}} = \frac{{{V_1}}}{{{V_2}}} \cr & \Rightarrow {\lambda _{{m_2}}} = \frac{{6.22 \times {{10}^3}}}{{{{10}^4}}} = 0.622\mathop {\text{A}}\limits^ \circ \cr} $$

Wavelength of emitted photon from $${n^{th}}$$ state to the ground state, $$\frac{1}{{{\Lambda _n}}} = R{Z^2}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{n^2}}}} \right)$$
$${\Lambda _n} = \frac{1}{{R{Z^2}}}{\left( {1 - \frac{1}{{{n^2}}}} \right)^{ - 1}}$$
Since $$n$$ is very large, using binomial theorem
$$\eqalign{ & {\Lambda _n} = \frac{1}{{R{Z^2}}}\left( {1 + \frac{1}{{{n^2}}}} \right) \cr & {\Lambda _n} = \frac{1}{{R{Z^2}}} + \frac{1}{{R{Z^2}}}\left( {\frac{1}{{{n^2}}}} \right) \cr & {\text{As we know, }}{\lambda _n} = \frac{{2\pi r}}{n} = 2\pi \left( {\frac{{{n^2}{h^2}}}{{4{\pi ^2}mZ{e^2}}}} \right)\frac{1}{n} \propto n \cr & {\Lambda _n} \approx A + \frac{B}{{\lambda _n^2}} \cr} $$

$$\eqalign{ & \phi = 6.2\,eV = 6.2 \times 1.6 \times {10^{ - 19}}J \cr & V = 5\,volt \cr & \frac{{hc}}{\lambda } - \phi = e{V_0} \cr & \Rightarrow \lambda = \frac{{hc}}{{\phi + e{V_0}}} \cr & = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}}\left( {6.2 + 5} \right)}} \cr & \approx {10^{ - 7}}m \cr} $$

Power of radio transmitter $$= 10 kW$$
$$ = 10000\,W$$
Operating frequency of transmitter $$ = 880\,kHz = 880 \times {10^3}Hz$$
Number of photons emitted per second
$$\eqalign{ & n = \frac{P}{E} = \frac{P}{{h\nu }} \cr & = \frac{{{{10}^4}}}{{6.63 \times {{10}^{ - 34}} \times 880 \times {{10}^3}}}\,\,\left[ {h = 6.63 \times {{10}^{ - 34}}J{\text{ - }}s} \right] \cr & = 1.72 \times {10^{31}} \cr} $$

Let $$m$$ be the mass of each particle, then $${\lambda _1} = \left( {\frac{h}{{m{v_1}}}} \right)$$   and $${\lambda _2} = \left( {\frac{h}{{m{v_2}}}} \right),$$    where $${{v_1}}$$ and $${{v_2}}$$ are the velocities of two particles as shown in the figure.
$${{\vec v}_{CM}} = \frac{{m{{\vec v}_1} + m{{\vec v}_2}}}{{2m}} = \frac{{{{\vec v}_1} + {{\vec v}_2}}}{2}$$
Velocity of $$A$$ w.r.t. $$C$$ frame is
$$\eqalign{ & {{\vec v}_{1c}} = {{\vec v}_1} - {{\vec v}_{CM}} = \frac{{{{\vec v}_1} - {{\vec v}_2}}}{2} \cr & \left| {{{\vec v}_{1c}}} \right| = \frac{{\sqrt {{{\vec v}_1} - {{\vec v}_2}} }}{2} = \left| {{{\vec v}_{2c}}} \right| \cr} $$
So, required wavelength is
$$\eqalign{ & \lambda = \frac{h}{{m\left| {{{\vec v}_{1c}}} \right|}} = \frac{h}{m} \times \frac{2}{{\frac{h}{m}\sqrt {\frac{1}{{\lambda _2^1}} + \frac{1}{{\lambda _2^2}}} }} \cr & = \frac{{2{\lambda _1}{\lambda _2}}}{{\sqrt {\lambda _1^2 + \lambda _2^2} }} \cr} $$

From photoelectric equation
$$\eqalign{ & h\nu = W + e{V_0}\,\,\left( {{\text{where,}}\,W = {\text{work}}\,{\text{function}}} \right) \cr & {\text{So}}\,\,\frac{{hC}}{\lambda } = W + 3e{V_0}\,.......\left( {\text{i}} \right) \cr & {\text{Also,}}\,\,\frac{{hC}}{{2\lambda }} = W + e{V_0} \cr & \Rightarrow \frac{{hC}}{\lambda } = 2W + 2e{V_0}\,.......\left( {{\text{ii}}} \right) \cr} $$
Subtracting Eq. (i) from Eq. (ii), we get
$$0 = W - e{V_0} \Rightarrow W = e{V_0}$$
From Eq. (i),
$$\frac{{hc}}{\lambda } = e{V_0} + 3e{V_0} = 4e{V_0}$$
The threshold wavelength is given by
$${\lambda _{{\text{th}}}} = \frac{{hc}}{W} = \frac{{4e{V_0}\lambda }}{{e{V_0}}} = 4\lambda $$

According to de-Broglie relation, wavelength of a particle is given by $$\lambda = \frac{h}{p}$$
where, $$h$$ is Planck's constant and wavelength of an electron is given by
$$\eqalign{ & {\lambda _e} = \frac{h}{{{p_e}}} \cr & {\text{but}}\,\lambda = {\lambda _e},{\text{ so }}p = {p_e} \cr & {\text{or}}\,\,{m_e}{v_e} = mv \cr & {\text{or}}\,\,v = \frac{{{m_e}{v_e}}}{m} \cr & {\text{Here,}}\,\,{m_e} = 9.1 \times {10^{ - 31}}\;kg \cr & {v_e} = 3 \times {10^6}\;m{s^{ - 1}} \cr & {\text{and}}\,\,m = 1mg \cr & = 1 \times {10^{ - 6}}\,kg \cr & \therefore v = \frac{{9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^6}}}{{1 \times {{10}^{ - 6}}}} \cr & = 2.7 \times {10^{ - 18}}\;m/s \cr} $$

Total number of ions in the rod
$$\eqalign{ & N = \left( {2 \times {{10}^{25}}} \right) \times \left[ {\frac{\pi }{4} \times {{\left( {{{10}^{ - 2}}} \right)}^2}} \right] \times \left( {5 \times {{10}^{ - 2}}} \right) \cr & = 7.85 \times {10^{19}} \cr} $$
The energy of excitation
$$\eqalign{ & E = N \times \frac{{hc}}{\lambda } \cr & = \left( {7.85 \times {{10}^{19}}} \right) \times \frac{{\left( {6.6 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{6.6 \times {{10}^{ - 7}}}} \cr & = 23.55\,J \cr} $$
Average power $$P = \frac{E}{t} = \frac{{23.55}}{{{{10}^{ - 7}}}} = 23.55 \times {10^7}W = 235\,MW$$