Dual Nature of Matter and Radiation MCQ Questions & Answers in Modern Physics | Physics
Learn Dual Nature of Matter and Radiation MCQ questions & answers in Modern Physics are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
81.
The ratio of the $${\lambda _{\min }}$$ in a Coolidge tube to $${\lambda _{{\text{de Broglie}}}}$$ of the electrons striking the target depends on accelerating potential $$V$$ as
A
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \sqrt V $$
B
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto V$$
C
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \frac{1}{{\sqrt V }}$$
D
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \frac{1}{V}.$$
82.
The $$21\,cm$$ radiowave emitted by hydrogen in interstelar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly
A
$${10^{ - 17}}J$$
B
$$1\,J$$
C
$$7 \times {10^{ - 6}}J$$
D
$${10^{ - 24}}J$$
Answer :
$${10^{ - 24}}J$$
The energy of emitted photon is given by
$$E = \frac{{hc}}{\lambda }$$
Given, wavelength, $$\lambda = 21\,cm = 0.21\,m$$
So, $$E = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.21}}$$
$$ = {10^{ - 24}}\,J$$
83.
$$1.5\,mW$$ of $$400\,nm$$ light is directed at a photoelectric cell. If 0.10 per cent of the incident photons produce photoelectrons, then find the current in the cell.
A
$$4.8\,\mu A$$
B
$$48\,\mu A$$
C
$$1.8\,\mu A$$
D
$$0.48\,\mu A$$
Answer :
$$0.48\,\mu A$$
The number of photons per second directed at the cell is
$$\eqalign{
& n = \frac{{{\text{Power}}}}{{hv}} = \frac{{P\lambda }}{{hc}} = \frac{{1.5 \times {{10}^{ - 3}} \times 400 \times {{10}^{ - 9}}}}{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \cr
& n = \frac{1}{{3.31}} \times {10^{16}}\left( {{\text{photons/s}}} \right) \cr} $$
But it is given that only $$0.1\% $$ of these photons produce photoelectrons. Therefore number of photoelectrons produced per second is
$${n_e} = \frac{{0.1}}{{100}} \times \frac{{{{10}^{16}}}}{{3.31}} = \frac{{{{10}^{13}}}}{{3.31}}$$
Therefore, the current is
$$\eqalign{
& I = {n_e}e \cr
& = \frac{{{{10}^{13}} \times 1.6 \times {{10}^{ - 19}}}}{{3.31}} = 0.48 \times {10^{ - 6}} = 0.48\,\mu A \cr} $$
84.
An electron of mass $$m$$ and a photon have same energy $$E.$$ The ratio of de-Broglie wavelengths associated with them is
($$c$$ being velocity of light)
A
$${\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$$
B
$$c{\left( {2mE} \right)^{\frac{1}{2}}}$$
C
$$\frac{1}{c}{\left( {\frac{{2m}}{E}} \right)^{\frac{1}{2}}}$$
D
$$\frac{1}{c}{\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$$
Since, it is given that electron has mass $$m.$$ de-Broglie’s wavelength for an electron will be given as
$${\lambda _e} = \frac{h}{P}\,......\left( {\text{i}} \right)$$
where, $$h$$ = Planck’s constant
$$P$$ = Linear momentum of electron
As kinetic energy of electron
$$E = \frac{{{P^2}}}{{2m}} \Rightarrow P = \sqrt {2mE} \,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$${\lambda _e} = \frac{h}{{\sqrt {2mE} }}\,......\left( {{\text{iii}}} \right)$$
Energy of a photon can be given as
$$\eqalign{
& E = hv \cr
& \Rightarrow E = \frac{{hc}}{{{\lambda _p}}} \cr
& \Rightarrow {\lambda _p} = \frac{{hc}}{E}\,......\left( {{\text{iv}}} \right) \cr} $$
Hence, $${\lambda _p} = $$ de-Broglie's wavelength of photon. Now, divide Eq. (iii) by Eq. (iv), we get
$$\eqalign{
& \frac{{{\lambda _e}}}{{{\lambda _p}}} = \frac{h}{{\sqrt {2mE} }} \cdot \frac{E}{{hc}} \cr
& \Rightarrow \frac{{{\lambda _c}}}{{{\lambda _p}}} = \frac{1}{c} \cdot \sqrt {\frac{E}{{2m}}} \cr} $$
85.
A source $${S_1}$$ is producing, $${10^{15}}$$ photons/s of wavelength $$5000\,\mathop {\text{A}}\limits^ \circ .$$ Another source $${S_2}$$ is producing $$1.02 \times {10^{15}}$$ photons per second of wavelength $$5100\,\mathop {\text{A}}\limits^ \circ .$$ Then, $$\frac{{\left( {{\text{power}}\,{\text{of}}\,{S_2}} \right)}}{{\left( {{\text{power}}\,{\text{of}}\,{S_1}} \right)}}$$ is equal to
A
1.00
B
1.02
C
1.04
D
0.98
Answer :
1.00
Number of photons emitted per second is given by
$$\eqalign{
& n = \frac{P}{{\left( {\frac{{hc}}{\lambda }} \right)}}\,\,\,\left[ {_{\frac{{hc}}{\lambda }\, = \,{\text{Energy}}}^{P\, = \,{\text{Power}}}} \right] \cr
& {\text{So,}}\,\,P = \frac{{nhc}}{\lambda } \cr} $$
So, for two different situations,
$$ \Rightarrow \frac{{{P_2}}}{{{P_1}}} = \frac{{{n_2}{\lambda _1}}}{{{n_1}{\lambda _2}}} = \frac{{1.02 \times {{10}^{15}} \times 5000}}{{{{10}^{15}} \times 5100}} = 1$$
86.
An X-ray tube is operated at $$15\,kV.$$ Calculate the upper limit of the speed of the electrons striking the target.
A
$$7.26 \times {10^7}\,m/s$$
B
$$7.62 \times {10^7}\,m/s$$
C
$$7.62 \times {10^7}\,cm/s$$
D
$$7.26 \times {10^9}\,m/s$$
Answer :
$$7.26 \times {10^7}\,m/s$$
The maximum kinetic energy of an electron accelerated through a potential difference of $$V$$ volt is $$\frac{1}{2}m{v^2} = eV$$
∴ maximum velocity $$v = \sqrt {\frac{{2eV}}{m}} $$
$$\eqalign{
& v = \sqrt {\frac{{2 \times 1.6 \times {{10}^{ - 19}} \times 15000}}{{9.1 \times {{10}^{ - 31}}}}} , \cr
& v = 7.26 \times {10^7}\,m/s \cr} $$
87.
The threshold wavelength of the tungsten is $$2300\,\mathop {\text{A}}\limits^ \circ .$$ If ultraviolet light of wavelength $$1800\,\mathop {\text{A}}\limits^ \circ $$ is incident on it, then the maximum kinetic energy of photoelectrons would be about -
88.
If the momentum of electron is changed by $$P,$$ then the de Broglie wavelength associated with it changes by $$0.5\% .$$ The initial momentum of electron will be:
A
$$200\,P$$
B
$$400\,P$$
C
$$\frac{P}{{200}}$$
D
$$100\,P$$
Answer :
$$200\,P$$
The de-Broglie’s wavelength associated with the moving electron $$\lambda = \frac{h}{P}$$
Now, according to problem
$$\eqalign{
& \frac{{d\lambda }}{\lambda } = - \frac{{dp}}{P}; \cr
& \frac{{0.5}}{{100}} = \frac{P}{{P'}} \cr
& \Rightarrow P' = 200\,P \cr} $$
89.
According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal Versus the frequency, of the incident radiation gives a straight line whose slope
A
depends both on the intensity of the radiation and the metal used
B
depends on the intensity of the radiation
C
depends on the nature of the metal used
D
is the same for the all metals and independent of the intensity of the radiation
Answer :
is the same for the all metals and independent of the intensity of the radiation
From Equation $$K.E. = hv - \phi $$
slope of graph of $$K.E\,\& \,v$$ is $$h$$ (Plank's constant)
which is same for all metals
90.
The short wavelength limit of continuous X-radiation emitted by an X-ray tube operating at $$30\,kV$$ is $$0.414\,\mathop {\text{A}}\limits^ \circ .$$ Calculate Planck's constant.