MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 2

A microscopic particle of mass

$10^{-12}$ kg is moving with a velocity of

$10^{2}$ m/s. Then the de-Broglie wavelength associated with the particle is

0%
$6.6\times 10^{-24}m$
0%
$6.6\times 10^{34}m$
0%
$3.3\times 10^{24}m$
0%
$0$

Explanation

$\lambda =\dfrac{h}{m\nu }$

$=\dfrac{6.6\times 10^{-34}}{10^{-12}\times 10^{2}}$

$=6.6\times 10^{-24}m$

So, the answer is option (A).

The momentum of a photon of frequency v is

0%
$\dfrac{hv}{c}$
0%
$hvc$
0%
$\dfrac{h}{cv}$
0%
$\dfrac{v}{ch}$

Explanation

Hint:

Energy mass relation for photon is given as,

$E = m{c^2}$

Where $E$ is the total energy of the photon, $m$ is the moving mass of the photon, and $c$ is the speed of light

Correct Option is A.

Explanation for correct answer:

$\bullet$ A photon is moving with frequency $v$ , then its energy is given as,

$E = hv$

$\bullet$ By energy mas relation, we have

$E = m{c^2}$

$\Rightarrow m{c^2} = hv$

$\Rightarrow mc = \dfrac{{hv}}{c}$
$\Rightarrow p = \dfrac{{hv}}{c}$

Where $p$ is the momentum of the photon.

This is the required expression. Momentum of photon is given by, $p = \dfrac{{hv}}{c}$

Thus, Option A is correct.

The momentum of a photon of wavelength 0.01

$A^{0}$ is

0%
$6.6\times 10^{-32}kg$ m/s
0%
$6.6\times 10^{-20}kg$ m/s
0%
$6.6\times 10^{-22}kg$ m/s
0%
$6.6\times 10^{-12}kg$ m/s

Explanation

de Broglie wavelength

$\lambda =\dfrac{h}{p}$

Thus momentum of the photon

$p=\dfrac{h}{\lambda }$

where wavelength of the photon

$\lambda = 0.01 \times 10^{-10} m$

$p=\dfrac{6.6\times 10^{-34}}{0.01\times 10^{-10}}$

$p= 6.6\times 10^{^{-22}}kg\ m/s$

So, the answer is option (C).

An electron accelerated under a p.d. of V volt has a certain wavelength

$\lambda$ . Mass of the proton is 2000 times the mass of an electron. If the proton has to have the same wavelength

$\lambda$ , then it will have to be accelerated under p.d. of (volts):

0%
$100$
0%
$2000$
0%
$V/2000$
0%
$\sqrt{2000}$

Explanation

$\dfrac{1}{2}mv^{2}=eV$

$mv=\ \sqrt{2meV}$

$Now \ \lambda _{e}=\dfrac{h}{mv}$

$=\dfrac{h}{\sqrt{2meV}}$

$\lambda_p =\ \dfrac{h}{\sqrt{2\times 2000m\times e\times V^{1}}}$

$Given,\ \lambda _{e} = \lambda _p$

$V^{1} = \dfrac{V}{2000}$

The value of de Broglie wavelength of an electron moving with a speed of

$6.6 \times 10^{5} ms^{-1}$ is approximately

0%
11 $A^{0}$
0%
111 $A^{0}$
0%
211 $A^{0}$
0%
311 $A^{0}$

Explanation

$\lambda =\dfrac{h}{m\nu }$

$=\dfrac{6.6\times 10^{-34}}{9.1\times 10^{-31}\times 6.6\times 10^{5}}$

$=11\ A^{\circ}$

So, the answer is option (A).

The de Broglie wavelength associated with a particle at rest is

0%
0
0%
$\infty$
0%
finite
0%
data insufficient

Explanation

The relationship between de Broglie wavelength

$(\lambda)$ and the momentum

$(p)$ of a photon is given by:

$\lambda=\dfrac{h}{p}$ . So, the de Broglie wavelength associated with a particle at rest

$(p=0)$ is infinity.

So, the answer is option (B).

Light of wavelength

$5000\overset { 0 }{ A }$ is incident on a sensitive surface. If the surface has received

${ 10 }^{ -7 }J$ of energy, then number of photon just falling on the surface are

0%
$2.5\times 10^6$
0%
$5\times 10^6$
0%
$2.5\times{ 10 }^{ 11 }$
0%
$5\times{ 10 }^{ 11 }$

Explanation

energy of 1 proton =

$\frac{hc}{\lambda }$
=

$\frac{hc}{5000}$

$E_1=\frac{12400}{5000}eV$

$n\times E_1=10^{-7}$

$n=\frac{10^{-7}\times 5000}{12400\times 1.6\times 10^{-19}}$
=

$2.5\times 10^{11}$

Energy of a photon of light of wavelength 450 nm is

0%
$4.4\times 10^{-19}J$
0%
$2.5\times 10^{-19}J$
0%
$1.25\times 10^{-17}J$
0%
$2.5\times 10^{-17}J$

Explanation

As we know,

$E=\dfrac{hc}{\lambda}=\dfrac{6.62\times 10^{-34}\times 3\times 10^8}{450\times 10^{-9}}=4.4\times 10^{-19}J$

Two different sources of light A and B have wavelength

$0.7\mu m$ and

$0.3\mu m$ respectively. Then which of the following statement is true.

0%
A has greater energy than B
0%
B has greater energy than A
0%
Both has equal energy
0%
None of the above

Explanation

$\lambda_A=0.7\times 10^{-6}m$ ,

$\lambda_B=0.3\times 10^{-1}m$

$E=\dfrac {hc}{\lambda}$

$\lambda_A > \lambda_B$
Hence, E_A < E_B

The ratio of energy of photon of

$\lambda$ = 2000 A

$^{o}$ to that of

$\lambda$ = 4000 A

$^{o}$ is :

0%
2
0%
1/4
0%
4
0%
1/2

Explanation

$_1$ = hv

$_1$ = h

$\displaystyle\frac{c}{\lambda_1}$ =

$\displaystyle\frac{hc}{2000}$
E

$_2$ = hv

$_2$ = h

$\displaystyle\frac{c}{\lambda_2}$ =

$\displaystyle\frac{hc}{4000}$

$\displaystyle\frac{E_1}{E_2}$ =

$\displaystyle\frac{hc \times 4000}{2000 \times hc}$ = 2

For the same kinetic energy, the momentum shall be minimum for

0%
Electron
0%
Proton
0%
Neutron
0%
Alpha particle

Explanation

$\because$ Kinetic energy,

$K=\displaystyle\frac{P^2}{2m}$
Here

$K$ is constant, so

$P\;\propto\;\sqrt{m}$
Hence electron has minimum momentum

Frequency of a photon having energy

$66\ eV$ is

0%
$8\times 10^{-15}Hz$
0%
$12\times 10^{-15}Hz$
0%
$16\times 10^{15}Hz$
0%
None of these

Explanation

As we know,

$E=hv \Rightarrow v=\dfrac{E}{h}=\dfrac{66\times 1.6\times 10^{-19}}{6.6\times 10^{-34}}=16\times 10^{15}Hz$

A device for producing intense and coherent (waves in step) beam of light is called a:

0%
laser
0%
maser
0%
search light
0%
radar

Louis de -Broglie is credited for his work on

0%
Theory of relativity
0%
Electromagnetic theory
0%
Matter waves
0%
Law of distribution of velocities

When light falls on certain materials, electrons are released. These electrons are called :

0%
extra electrons
0%
valency electrons
0%
free electrons
0%
photoelectrons

Einstein got his Nobel Prize for

0%
His theory of relativity
0%
Explanation of Photoelectric effect
0%
His theory of atomic heats of solids
0%
None of the above

A photon of light from which of the following electromagnetic radiations carries the greater amount of energy?

0%
Blue
0%
Green
0%
Orange
0%
Red
0%
Yellow

Explanation

We know that energy of a photon is given by ,

$E=h\nu$

where

$\nu$ is the frequency of radiation ,

now in electromagnetic visible spectrum we have decreasing order of frequency as ,

$VIBGYOR$ , since energy is directly proportional to frequency as shown by above relation, we can see by

$VIBGYOR$ that blue radiation has greater frequency in the given radiations , therefore it has greater amount of energy.

The frequency of radio waves corresponding to a wavelength of

$10\ m$ is

0%
$3\times 10^{7} Hz$
0%
$3.3\times 10^{8} Hz$
0%
$3\times 10^{9} Hz$
0%
$3\times 10^{-7} Hz$

Explanation

Wavelength,

$\lambda = 10 \: m$

Using the relation

$\nu=\dfrac{c}{ \lambda }= \dfrac{ 3 \times 10^8}{10}= 3 \times 10^7\: Hz$

The emission of photon without using an external agency for its emission in a system is called __________.

0%
light amplification
0%
induced absorption
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stimulated emission
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spontaneous emission

The de-Broglie wavelength

$\lambda$ of a particle

0%
is proportional to mass.
0%
is proportional to impulse.
0%
is inversely proportional to impulse.
0%
does not depend on impulse.

Explanation

De-Broglei wavelength,

$\lambda = \dfrac{h}{p}$ where,

$h$ is Planck's constant

Impulse,

$I = \Delta p = p$

$\therefore$

$\lambda = \dfrac{h}{I}$

$\implies$

$\lambda \propto \dfrac{1}{I}$

Thus, De-Broglei wavelength is inversely proportional to the impulse.

Also,

$p = mv$

$\therefore$

$\lambda = \dfrac{h}{mv}$

$\implies \lambda \propto \dfrac{1}{m}$

LASER beams are used to measure long distances because

0%
they are monochromatic.
0%
they are coherent.
0%
they are highly polarised.
0%
they have high degree of parallelism.

When is the energy received from the photon maximum, if a photon strikes an electron?

0%
The electron does not spin when it is struck
0%
The photon has a high velocity
0%
The photon has a low velocity
0%
The photon has a long wavelength
0%
The photon has a high frequency

Explanation

Energy of photon

$E = h\nu$

$\therefore$

$E \propto \nu$

Thus the energy of photon is maximum when it has high frequency.

The de-Broglie wavelength of electron in second Bohr's orbit is equal to ____________.

0%
circumference of the orbit
0%
half the circumference of the orbit
0%
twice the circumference of the orbit
0%
four times circumference of the orbit

Explanation

Conservation of angular momentum gives:

$mvr=\dfrac { nh }{ 2\pi } \\ \Rightarrow \dfrac { h }{ mv } =\dfrac { 2\pi r }{ n }$

De Broglie's wavelength,

$\lambda =\dfrac { h }{ mv }$

Thus, for

$n =2$

$\Rightarrow \lambda =\dfrac { 2\pi r }{ 2 } =\dfrac { circumference }{ 2 }$

Which of the following statements about photon is incorrect?

0%
Photon exerts no pressure
0%
Rest mass of photon is zero
0%
None of these
0%
Energy of photon is $hv$

Explanation

The photon moves with a velocity of light and has energy

$hv$ . Therefore, it exerts pressure

$P$

$P=\dfrac{I}{C}$

When light falls on certain materials, electrons are released.These electrons are called :

0%
Extra electrons
0%
Valency electrons
0%
Free electrons
0%
Photoelectrons

According to de Broglie, Which of the following statements is true about the wavelength of a moving particle?

0%
It is never large enough to measure.
0%
It is proportional to the speed of the particle.
0%
It is inversely proportional to the momentum of the particle.
0%
It is equal to Planck's constant
0%
It has no effect on the behavior of electrons.

Explanation

de Broglie wavelength associated with the particle

$\lambda = \dfrac{h}{p}$

$\implies$

$\lambda \propto \dfrac{1}{p}$ where

$h$ is the Planck's constant

Thus de Broglie wavelength is inversely proportional to the momentum of the particle

Identify which of the above graph best describe the relationship between the energy of the photon and its wavelength ?

0%
Graph A
0%
Graph B
0%
Graph C
0%
Graph D
0%
Graph E

Explanation

Energy of a photon

$E = \dfrac{hc}{\lambda}$ where

$\lambda$ is the wavelength of the photon

$\implies$

$E\lambda = hc = constant$

Thus E Vs

$\lambda$ graph is a hyperbola.

Hence option B is correct.

Relation between wavelength of photon and electron of same energy is

0%
$\lambda_{ph} > \lambda_e$
0%
$\lambda_{ph} < \lambda_e$
0%
$\lambda_{ph} = \lambda_e$
0%
$\frac{\lambda_e}{\lambda_{ph}}=constant$

Explanation

For the same energy, the momentum of electron is more than that of photon

$\Rightarrow { p }_{ e }>{ p }_{ ph }$

${ \lambda }_{ e }=\dfrac { h }{ { p }_{ e } } \quad \quad and\quad \quad { \lambda }_{ ph }=\dfrac { h }{ { p }_{ ph } } \\ \Rightarrow { \lambda }_{ ph }>{ \lambda }_{ e }$

Which of the following graphs represents the energy of a photon vs. its frequency?

Identify the electromagnetic wave whose photons has the highest energy .

0%
X-ray
0%
ultraviolet light
0%
green light
0%
microwave
0%
radio wave

Explanation

Energy of photon

$E = h\nu$

$\implies E \propto \nu$

Increasing order of frequency :

Radio wave < micro wave < green light < ultraviolet light < X-ray.

Thus frequency of X-ray is the highest.

$\implies$ Energy of X-ray photon is the highest.

Identify the reason why an observer cannot detect the wave nature of a fast moving truck .

0%
The momentum of the truck is too large
0%
The velocity of the wave is too large
0%
There are no waves to be detected
0%
The frequency is too low
0%
The wavelengths are too small

Explanation

The de-Broglie's wavelength associated with a moving matter is given by ,

$\lambda=h/p$ ,

where

$h=$ planck's constant ,

$p=$ momentum of matter ,

now , a moving truck has a very small momentum therefore the wavelength associated with it is very large or frequency is very much low , so an observer cannot observe it .

An electron has a wavelength of 1 nm. Then, the kinetic energy of the electron is

0%
$1.8 \times 10^{-6} \, J$
0%
$19.86 \times 10^{-17} \, J$
0%
$16.82 \times 10^{-19} \, J$
0%
$22.3 \times 10^{-9} \, J$

Explanation

From the equation

$E=\dfrac{hc}{\lambda }$
We have,

$h= 6.63 \times 10^{-34}, c= 3\times 10^8 m/s$

$\lambda =1nm = 10^{-9}m$

$E=\dfrac{6.63\times 10^{-34}\times3\times10^8}{10^{-9}}= 10^{-9}$

$E=19.86\times 10^{-17}\, J$

A laser beam is used for carrying out surgery because it

0%
Is highly coherent
0%
Is highly directional
0%
Can be sharply focussed
0%
Is highly monochromatic

Max K.E of photo electrons depend upon light's

0%
intensity
0%
frequency
0%
wavelength
0%
all of these

Explanation

Maximum kinetic energy

$k.E_{max} = h\nu_{incident} - \phi$
where

$\nu_{incident}$ is the frequency of the incident light and

$\phi$ is the work function of the metal surface.
Thus maximum kinetic energy of the photo electrons depends upon light's wavelength.

The energy of a photon varies directly with its

0%
frequency
0%
wavelength
0%
speed
0%
rest mass

Explanation

Energy of photon

$E = h\nu$
Thus, energy of the photon varies with the frequency of the incident photon.

The de-Broglie wavelength associated with thermal neutron ( t=51C say) is of the order of the

0%
distance between the atoms in the crystal
0%
size of the nucleus
0%
Bohr's radius
0%
size of grain

Explanation

$T = 51^\circ C$

So, temp in Kelvin =

$51 + 273 = 324 K$

de-broglie wavelength

$\lambda = \dfrac{h}{p}$

$p = \sqrt{2mE}$

$\lambda = \dfrac{h}{\sqrt{2mE}}$

$m$ of neutron

$= 1.67 \times 10^{-27} kg$

Also,

$E = \dfrac{3}{2} kT$

where

$k = \text{Boltzman constant} = 1.38 \times 10^{-23} J \, mol^{-1} \, K^{-1}$

So,

$\lambda = \dfrac{h}{\sqrt{3mkT}}$

Substituting the values,

$\lambda = 1.39 \times 10^{10} m$

which is the order of distance between atoms in the crystal.

If the wave properties of a particle are difficult to observe, it is probably due to the particle's

0%
small size
0%
large mass
0%
low momentum
0%
high charge

De-Broglie wavelength of an atom at absolute temperature

$T\ K$ will be

0%
$\dfrac {h}{\sqrt {3mKT}}$
0%
$\dfrac {h}{mKT}$
0%
$\dfrac {\sqrt {2mKT}}{h}$
0%
$\sqrt {2mKT}$

Explanation

Key Concept The average kinetic energy of atom at temperature

$T\ K$ is

$E = \dfrac {3}{2}KT$
Now, kinetic energy of an atom

$E = \dfrac {P^{2}}{2m}$

$\therefore P = \sqrt {2ME}$

$\therefore$ de-Broglie wavelength

$\lambda = \dfrac {h}{P} = \dfrac {h}{\sqrt {2mE}}$

$\Rightarrow \lambda = \dfrac {h}{\sqrt {2m\times \dfrac {3}{2}KT}} = \dfrac {h}{\sqrt {3mKT}}$ .

Which color of light has the greatest energy per photon?

0%
blue
0%
green
0%
violet
0%
red

Explanation

Energy of photon

$E = h\nu$
Thus more is the frequency of light, more is the energy.
Since, violet light has the highest frequency. So violet light has the highest energy.

The work function of

$Na$ metal is

$2.3\ eV$ , then the threshold wavelength lies in the following region of

$EM$ spectrum

0%
ultraviolet
0%
X-ray
0%
violet
0%
yellow

Explanation

$\phi=\dfrac{12400}{\lambda}$ putting

$\phi=2.3eV$ we get

$\lambda=12400/2.3=5391 Angstrom$ which is for

$yellow$ colour.

Option D is correct.

If the mass of a microscopic particle a well as its speed are halved, the de-broglie wavelength associated with the particle will

0%
increased by a factor more than $2$
0%
increases by a factor of $2$
0%
decreases by a factor of $2$
0%
decrease by a factor more than $2$

Explanation

$\lambda$ is directly proportional to $\dfrac{1}{mv}$

$\dfrac{ \lambda_1}{ \lambda_2} = \dfrac{m_2 v_2}{ m_1 v_1}$

$\dfrac{ \lambda_1}{ \lambda_2} = \dfrac{1}{2} \times \dfrac{1}{2}$

$\lambda_2 = 4 \lambda_1$

If a proton and electron have the same de-Broglie wavelength, then

0%
Kinetic energy of electron < kinetic energy of proton
0%
Kinetic energy of electron = kinetic energy of proton
0%
Momentum of electron > momentum of proton
0%
Momentum of electron = momentum of proton
0%
Momentum of electron < momentum of proton

Explanation

de-Broglie wavelength is given by

$\lambda=\dfrac{h}{p}$ ,

Where

$h$ is

$Planck's$ constant so

$same$ for both electron and proton.

$p$ is

$momentum$ of the particle, as both have same wavelength so

$momentum$ should be

$same$ for both of them.

The de-Broglie wavelength associated with a particle with rest mass

$m_0$ and moving with speed of light in vacuum is ___________.

0%
$\dfrac{h}{m_0c}$
0%
$0$
0%
$\infty$
0%
$\dfrac{m_0c}{h}$

Explanation

De-Broglie wavelength

$\lambda=\dfrac{h}{p}$

where

$h=$ Plank's constant

$p=$ momentum associated with a particle

Given:

$mass=m_o$

$speed=c$

Therefore

$p=m_oc$

Hence

$\lambda=\dfrac{h}{p}=\dfrac{h}{m_oc}$

Therefore the correct option is (A).

The energy of a photon corresponding to the visible light of maximum wavelength is approximately

0%
1 eV
0%
1.6 eV
0%
3.2 eV
0%
7 eV

Explanation

$Maximum\quad wavelength\quad of\quad red\quad light\quad =750nm\\ =750\times { 10 }^{ -9 }m\\ Energy,\quad E=\frac { hc }{ \lambda } \\ =\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 750\times { 10 }^{ -9 } } \times \frac { 1 }{ 1.6\times { 10 }^{ -9 } } \\ =1.6eV$

Total number of protons in

$10 g$ of calcium
carbonate is

$(N_A = 6.023 \times 10^{23})$

0%
(a) $12.04 \times 10^{24}$
0%
(b) $4.06 \times 10^{24}$
0%
(c) $2.01 \times 10^{24}$
0%
(d) $3.24 \times 10^{24}$

A proton is accelerated to

$225V$ . Its de-Broglie wavelength is:

0%
$0.0019nm$
0%
$0.02nm$
0%
$0.003nm$
0%
$0.4nm$

Explanation

Energy of proton =

$\cfrac{p^2}{2m} = qV$

$\cfrac{p^2}{2m} = 225eV$

$p = \sqrt{2m \times 225 \times 1.6 \times 10^{-19}}$

$\therefore \lambda = \cfrac{h}{p}$

$= \cfrac{6.626 \times 10^{-34}}{\sqrt{2m \times 225 \times 1.6 \times 10^{-19}}}$

$= 0.19 \times 10^{-11} m$

$= 1.9 \times 10^{-12} m$

$= 1.9 pm$ or

$0.001 nm$

The ratio of the energy of a photon with

$\lambda =150\ nm$ to that with

$\lambda =300\ nm$ is

0%
$2$
0%
$1/4$
0%
$4$
0%
$1/2$

Explanation

As we know,

$E=\dfrac {hc}{\lambda} \Rightarrow \dfrac {E_1}{E_2}=\dfrac {300}{150}=\dfrac {2}{1}$

Energy of photon whose frequency is

$10^{12}MHz$ , will be

0%
$4.14\times 10^{3}keV$
0%
$4.14\times 10^{2}keV$
0%
$4.14\times 10^{3}MeV$
0%
$4.14\times 10^{3}eV$

Explanation

As we know,

$E(eV)=\dfrac{hv}{e}=\dfrac{6.0\times 10^{-34}\times 10^{12}\times 10^6}{1.6\times 10^{-19}}=4.14\times 10^3eV.$

Photon and electron are given energy

$\left( {{{10}^{ - 2}}J} \right).$ Wavelengths associated with photon and electeon are

${\lambda _{ph}}$ and

${\lambda _{el}},$ then correct statement will be

0%
${\lambda _{ph}} > {\lambda _{el}}$
0%
${\lambda _{ph}} < {\lambda _{el}}$
0%
${\lambda _{ph}} = {\lambda _{el}}$
0%
$\dfrac{{{\lambda _{el}}}}{{{\lambda _{ph}}}} = c$

Explanation

The wavelength of the photon will be greater than that of the electron because the mass of a photon is less than that of an electron.

$\Rightarrow \lambda_{ph} > \lambda_el$

A particle at rest decays into two particles of masses 2 m and
3 m having non-zero velocities. The ratio of DE-Imbroglio wavelength of the particle is

0%
$3 : 2$
0%
$1 : 1$
0%
$1 : 3$
0%
$\sqrt{2} : 1$

Explanation

$\lambda \propto {1 \over m}$

${{{\lambda _1}} \over {{\lambda _2}}} = {{{m_2}} \over {{m_1}}} = {{3m} \over {2m}}$

$= 3:2$

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 2 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 2 - MCQExams.com

0:0:3

Practice Class 12 Medical Physics Quiz Questions and Answers

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