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

$$\eqalign{ & {\lambda _{\min }} = \frac{{hc}}{E} \cr & \therefore {\lambda _{\min }} = \frac{{12400}}{{80 \times {{10}^3}}}\mathop {\text{A}}\limits^ \circ = 0.155\,\mathop {\text{A}}\limits^ \circ \cr} $$
Energy of incident electrons is greater than the ionization energy of electrons in $$K$$-shell, the $$K$$-shell electrons will be knocked off. Hence, characteristic X-ray spectrum will be obtained.

As we know that stopping potential gives the maximum $$KE$$  of ejected electrons, so
$$\eqalign{ & K{E_{\max }} = e{V_0} \cr & {\text{So,}}\,\,1.6 \times {10^{ - 19}} \times {V_0} = 0.5 \times 1.6 \times {10^{ - 19}} \cr & \Rightarrow {V_0} = 0.5\,V \cr} $$

The work function has no effect on photoelectric current so long as $$hv > {W_0}.$$   The photoelectric current is proportional to the intensity of incident light. Since there is no change in the intensity of light, hence $${I_1} = {I_2}.$$

$$\eqalign{ & {\text{As we know that,}} \cr & \frac{1}{2}m{\left( {{V_{\max }}} \right)^2} = h\nu \cr & {\text{So,}}\,\,{V_{\max }} = \sqrt {\frac{{2h\nu }}{m}} \cr} $$

Acceleration, $$a = \frac{{Ee}}{m}.$$
Velocity of electron, $$v = at = \frac{{Eet}}{m}.$$
Wavelength, $$\lambda = \frac{h}{{mv}} = \frac{h}{{Eet}}.$$
Now $$\frac{{d\lambda }}{{dt}} = - \frac{h}{{eE{t^2}}}.$$

de-Broglie wavelength associated with a moving particle can be given as
$$\lambda = \frac{h}{p} = \frac{{2h}}{{\sqrt {2m\left( {KE} \right)} }}$$
At thermal equilibrium, temperature of neutron and heavy water will be same.
This common temperature is given as, $$T.$$
Also, we know that, kinetic energy of a particle
$$KE = \frac{{{p^2}}}{{2m}}$$
where, $$p =$$  momentum of the particle
$$m =$$  mass of the particle
Kinetic energy of the neutron is
$$KE = \frac{3}{2}kT$$
∴ De-Broglie wavelength of the neutron
$$\eqalign{ & \lambda = \frac{h}{p} = \frac{h}{{\sqrt {2m\left( {KE} \right)} }} \cr & = \frac{h}{{\sqrt {2m \times \frac{3}{2}kT} }} \cr & = \frac{h}{{\sqrt {3mkT} }} \cr} $$

We write Einstein's equation for the photoelectric effect,
$$\eqalign{ & hf - W = {E_K}\,......\left( {\text{i}} \right) \cr & 3hf - W = {{E'}_K}\,......\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii), we get,
$${{E'}_K} - 3{E_K} = 2W > 0.$$

According to de-Broglie, a moving particle sometimes acts as a wave and sometimes as a particle. The wave associated with moving particle is called matter wave or de-Broglie wave whose wavelength is called de-Broglie wavelength and it is given by
$$\lambda = \frac{h}{{mv}}$$
where, $$m$$ and $$v$$ are the mass and velocity of the particle and $$h$$ is Planck's constant.

Give that, only $$25\% $$  of $$200\,W$$  converter 2 electrical energy into light of yellow colour
$$\left( {\frac{{hc}}{\lambda }} \right) \times N = 200 \times \frac{{25}}{{100}}$$
Where $$N$$ is the No. of photons emitted per second,
$$h = $$  plank’s constant, $$c = $$  speed of light.
$$\eqalign{ & N = \frac{{200 \times 25}}{{100}} \times \frac{\lambda }{{hc}} \cr & = \frac{{200 \times 25 \times 0.6 \times {{10}^{ - 6}}}}{{100 \times 6.2 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \cr & = 1.5 \times {10^{20}} \cr} $$

The work function of the metal
$$\eqalign{ & {W_0} = h{f_0} = \left( {6.63 \times {{10}^{ - 34}}} \right) \times \left( {6 \times {{10}^{14}}} \right) \cr & = 3.98 \times {10^{19}}J \cr} $$
If $$f$$ is the frequency of the light, then
$$\eqalign{ & hf = {W_0} + e{V_0} \cr & \therefore f = \frac{{{W_0} + e{V_0}}}{h} \cr & = \frac{{3.98 \times {{10}^{ - 19}} + 1.6 \times {{10}^{ - 19}} \times 3}}{{6.63 \times {{10}^{ - 34}}}} \cr & = 1.32 \times {10^{15}}\,Hz \cr} $$