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

Formation of covalent bond is best explained by molecular orbital theory.

Cathode rays are negatively charged particles called as electrons
(A) Cathode rays possess very high kinetic energy due to their high velocity. When these highly energetic rays fall on platinum (a metal), their kinetic energy is converted to heat energy.
(B) Outside the discharge tube, if an electric field is applied, the cathode rays bend towards the positive plate.
(C) Cathode rays travel in straight lines. This can be proved by an arrangement which shows that cathode rays cast shadow of the object placed in straight line path of cathode rays.
(D) In certain substances like barium platinocyanides, zinc sulphate, diamond etc, they produce fluorescence. Thus, (B) is right option.

As work function $$W = h{v_0},$$  where $${v_0}$$ is the threshold frequency.
Greater the work function, greater is the threshold frequency. Therefore, the threshold frequency of sodium will be lesser than that for aluminium.

For de-Broglie wavelength, $$\lambda = \frac{h}{p}$$
For 1^st case, $${\lambda _1} = \frac{h}{p} = \frac{h}{{\sqrt {2mK} }}$$
For 2^nd case, $${\lambda _2} = \frac{h}{{\sqrt {2m\,16K} }} = \frac{h}{{4\sqrt {2mK} }} = \frac{{{\lambda _1}}}{4}$$
$${\lambda _2} = 25\% \,{\text{of}}\,{\lambda _1}$$
So, $$75\% $$  change in the wavelength takes place.

When electrons are acceleated through $$V$$ volt, the gain in $$KE$$  of the electron is given by
$$\eqalign{ & KE = \frac{1}{2}m{v^2} = eV \cr & {\text{Given,}}\,\,V = 100\,V \cr & \therefore KE = \left( {1.6 \times {{10}^{ - 19}}} \right) \times 100 = 1.6 \times {10^{ - 17}}J \cr} $$

Work function $$\phi $$ of metal $$= 2.28\,eV$$
Wavelength of light $$\lambda = 500\,nm = 500 \times {10^{ - 9}}m$$
$$\eqalign{ & K{E_{\max }} = \frac{{hc}}{\lambda } - \phi \cr & K{E_{\max }} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5 \times {{10}^{ - 7}} \times 1.6 \times {{10}^{ - 19}}}} - 2.28 \cr & K{E_{\max }} = 2.48 - 2.28 = 0.2\,ev \cr & {\lambda _{\min }} = \frac{h}{p} = \frac{h}{{\sqrt {2m{{\left( {KE} \right)}_{\max }}} }} \cr & {\lambda _{\min }} = \frac{{25}}{9} \times {10^{ - 9}} = 2.80 \times {10^{ - 9}}nm \cr & \therefore \lambda \geqslant 2.8 \times {10^{ - 9}}m \cr} $$

In X-ray tube, $${\lambda _{\min }} = \frac{{hc}}{{eV}}$$
In $${\lambda _{\min }} = \log \left( {\frac{{hc}}{e}} \right) - \log V$$
Clearly, $$\log {\lambda _{\min }}$$  versus $$\log \,V$$  graph slope is negative hence option (C) correctly depicts.

$$E = \frac{{hc}}{\lambda }$$
Number of photons emitted is
$$\eqalign{ & \frac{{Pt}}{{\left( {\frac{{hc}}{\lambda }} \right)}} = {n_0} \cr & {n_0} = \frac{{P\lambda t}}{{hc}} \cr} $$
Since the radiation is spherically symmetric, so total number of photons entering the sensor is $${n_0}$$ times the ratio of aperture area to the area of a sphere of radius $$\ell .$$
$$N = {n_0}\frac{{\pi {{\left( {2d} \right)}^2}}}{{4\pi {\ell ^2}}} = \frac{{P\lambda t}}{{hc}}\frac{{{d^2}}}{{{\ell ^2}}}$$

Final speed of $$\alpha $$ -particle will be less than the initial speed therefore,
$$\left( {\frac{h}{{{\lambda _f}}}} \right) < \left( {\frac{h}{{{\lambda _i}}}} \right)\,\,{\text{or}}\,\,{\lambda _f} > {\lambda _i}$$

$$\eqalign{ & E = eV = 2 \times {10^6} \times 1.6 \times {10^{ - 19}}J \cr & \Rightarrow \lambda = \frac{{hc}}{E} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3.2 \times {{10}^{ - 13}}}} \approx {10^{ - 2}}\mathop {\text{A}}\limits^ \circ \cr} $$