Either mechanical or electromagnetic waves

MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 1

A free particle with initial kinetic energy

$E$ and de-broglie wavelength

$\lambda$ enters a region in which it has potential energy

$U$ . What is the particle's new de-Broglie wavelength?

0%
$\lambda (1-U/E)^{-1/2}$
0%
$\lambda (1-U/E)$
0%
$\lambda (1-U/E)^{-1}$
0%
$\lambda (1-U/E)^{1/2}$

In which of the following photocell is not used?

0%
Burglar alarm
0%
Television camera
0%
Automatic street lights
0%
Vacuum cleaner

If the

$KE$ of a free electron doubles then its de-Broglie wavelength changes by a factor

0%
$\displaystyle\dfrac{1}{2}$
0%
$\displaystyle\dfrac{1}{\sqrt 2}$
0%
$\displaystyle2$
0%
$\displaystyle\sqrt 2$

Explanation

$\displaystyle \lambda =\dfrac{h}{p}=\dfrac{h}{\sqrt{2\left ( KE \right )m}}$

Thus when KE is doubled, the wavelength is changed by a factor of

$\dfrac{1}{\sqrt 2}$

The ratio of the energy of a photon of

$2000 \mathring { A }$ wavelength to that of

$4000\mathring { A }$ wavelength is :

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

Explanation

As we know,

$E = hc/\lambda$

So,

$E_1/E_2 = \lambda_2 /\lambda_1 = 4000/2000 = 2$

If a photon has velocity

$c$ and frequency

$v$ , then which of the following represents its wavelength

0%
$\dfrac {hc}{E}$
0%
$\dfrac {hv}{c}$
0%
$\dfrac {hv}{c^2}$
0%
$hv$

Explanation

If a photon has velocity c and frequency v, then

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

The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately:

0%
540 nm
0%
400 nm
0%
310 nm
0%
220 nm

Explanation

Minimum energy required for photoelectron emission is the work potential=

$4eV$

Energy of photon=

$\dfrac{hc}{\lambda}$

$\implies \dfrac{hc}{\lambda_{max}}=4eV$

$\lambda_{max}=\dfrac{(6.626\times 10^{-34})\times (3\times 10^8)}{4\times 1.6\times 10^{-19}}\approx 310nm$

Photon of frequency

$'v\ '$ has a momentum associated with it. If

$'c\ '$ is the velocity of light, the momentum is

0%
$v/c$
0%
$hvc$
0%
$hv/c$ $^{2}$
0%
$hv/c$

Explanation

We know that energy of photon is

$E=h\nu ....(1)$

Momentum of photo ,

$p=mc=mc^2/c=E/c$ where

$E=mc^2$

using (1),

$p=h\nu/c$

A Laser light of wavelength

$660 \,nm$ is used to weld Retina detachment. If a Laser pulse of width

$60\,ms$ and power

$0.5\,kW$ is used, the approximate number of photons in the pulse are:
[Take Planck's constant

$h=6.62\times 10^{-34}Js$ ]

0%
$10^{20}$
0%
$10^{18}$
0%
$10^{22}$
0%
$10^{19}$

Explanation

Power due to laser

$= \cfrac { \Delta n }{ \Delta t } .\cfrac { hc }{ \lambda }$

$\Delta n\rightarrow$ Number of photons per pulse

$\Delta t=60ms$

$0.5\times { 10 }^{ 3 }W=\cfrac { \Delta n }{ 60\times { 10 }^{ -3 } } \times \cfrac { 6.62\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 660\times 10^{ -9 } }$

$996.97\times { 10 }^{ 17 }=\Delta n$

$\Delta n=0.99\times { 10 }^{ 20 }$

$\Delta n\approx { 10 }^{ 20 }$

For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?

0%
An electron
0%
A proton
0%
An $\alpha-particle$
0%
A dust particle

If electron charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental constant h are taken as fundamental quantities, the permeability of vacuum

$\mu_0$ can be expressed in units of

0%
$\left (\dfrac {mc^2}{he^2}\right )$
0%
$\left (\dfrac {h}{me^2}\right )$
0%
$\left (\dfrac {hc}{me^2}\right )$
0%
$\left (\dfrac {h}{ce^2}\right )$

Explanation

The fine structure constant is defined as

$\alpha = \dfrac{e^2}{4\pi \epsilon _0 \hbar c}$ and is dimensionless where

$\hbar = \dfrac{h}{2\pi}$

$\therefore \alpha = \dfrac{e^2 c}{4\pi \epsilon _0 \hbar c^2}=\dfrac{e^2 c \epsilon _0 \mu _0}{4\pi \epsilon _0 \hbar}$ where

$c = \dfrac{1}{\sqrt{\epsilon _0 \mu _0}}$

$\Rightarrow \mu _0 = \dfrac{ \hbar}{e^2 c}$

If the de Broglie wavelengths associated with a proton and an

$\alpha -$ particle are equal, then the ratio of velocities of the proton and the

$\alpha -$ particle will be:

0%
$1:4$
0%
$1:2$
0%
$4:1$
0%
$2:1$

Explanation

Given de Broglie Wavelength

$\lambda_{p} = \lambda_{\alpha}$

So,

$\dfrac {h}{m_{p} \times v_{p}}$ =

$\dfrac {h}{m_{\alpha} \times v_{\alpha}}$

$\dfrac{v_{p}} {v_{\alpha}}$ =

$\dfrac {m_{\alpha}}{m_{p}}$ =

$\dfrac{4 m_{p}} {m_{p}}$

Because Mass of

$\alpha$ particle is 4 times mass of proton.

So 4:1.

Option C is correct answer

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to:

$(|e|=1.6 \times 10^{-19} C, m_e=9.1 \times 10^{-31}kg, h=6.6 \times 10^{-34} Js).$

0%
$0.5 \mathring {A}$
0%
$1.7 \mathring {A}$
0%
$2.4 \mathring {A}$
0%
$1.2 \mathring {A}$

Explanation

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

$\lambda = \dfrac{ 6.6 \times 10 ^ {-34} }{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 50}} = 1.72 \times 10^{-10} = 1.72 A$

Option B

The de-Broglie wavelength

$(\lambda_{B})$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state

$(\lambda_{G})$ by :

0%
$\lambda_{B} = \lambda_{G}/3$
0%
$\lambda_{B} = \lambda_{G}/2$
0%
$\lambda_{B} = 2\lambda_{G}$
0%
$\lambda_{B} = 3\lambda_{G}$

Explanation

de- Broglie wavelength

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

By conservation of angular momentum,

$mvr=\dfrac{nh}{2\pi}$

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

$\lambda=\dfrac{2\pi r}{n}$

$r=a_{0}\dfrac{n^{2}}{Z}$

$\lambda=\dfrac{2\pi a_{0}n}{Z}$

As the atom is hydrogen

$Z=1$

$\lambda_{B}=\dfrac{2\pi a_{0}3}{Z}$

$\lambda_{G}=\dfrac{2\pi a_{0}}{Z}$

$\lambda_{B}=3\lambda_{G}$

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are

${ \lambda }_{ 1 }$ and

${ \lambda }_{ 2 }$ , their de Broglie wavelength in the frame of reference attached to their centre of mass is:

0%
${ \lambda }_{ CM }={ \lambda }_{ 1 }={ \lambda }_{ 2 }\quad$
0%
$\cfrac { 1 }{ { \lambda }_{ CM } } =\cfrac { 1 }{ { \lambda }_{ 1 } } +\cfrac { 1 }{ { \lambda }_{ 2 } }$
0%
${ \lambda }_{ CM }=\cfrac { 2{ \lambda }_{ 1 }{ \lambda }_{ 2 } }{ \sqrt { { { \lambda }_{ 1 } }^{ 2 }+{ { \lambda }_{ 2 } }^{ 2 } } }$
0%
${ \lambda }_{ CM }=\left( \cfrac { { \lambda }_{ 1 }+{ \lambda }_{ 2 } }{ 2 } \right)$

Explanation

Since

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

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

Let

$v_1$ and

$v_2$ are the speeds of electrons

$v_{cm} = \dfrac{v_1 + v_2}{2}$

$\dfrac{h}{2m \lambda_{cm}} = \dfrac{1}{2}( \dfrac{h}{m \lambda_1} + \dfrac{h}{m \lambda_2})$

$\dfrac{1}{\lambda_{cm}} = \dfrac{1}{\lambda_1} + \dfrac{1}{\lambda_2}$

An electron (mass

$m$ ) with initial velocity

$\vec { v } ={ v }_{ 0 }\hat { i } +{ v }_{ 0 }\hat { j }$ is an electric field

$\vec { E } =-{ E }_{ 0 }\hat { k }$ . If

${ \lambda }_{ 0 }$ is initial de-Broglie wave length at time

$t$ is given by

0%
$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ 2{ m }^{ 2 }{ v }_{ 0 }^{ 2 } } } }$
0%
$\cfrac { { \lambda }_{ 0 }\sqrt { 2 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } }$
0%
$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 2+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } }$
0%
$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }_{ 0 }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } }$

Explanation

Initially

momentum,

$P=\dfrac{h}{\lambda_0}$

$\because V_i = \sqrt{V_0^2 + V_0^2} = \sqrt{2}V_0$

$m(\sqrt{2}V_0) = \dfrac{h}{\lambda_0}$

Now, velocity as a function of time

$= V_0\hat{i} + V_0\hat{j} + \dfrac{eE_0}{m} + \hat{k}$

because only force is acting in

$Z-dir^n$ . So velocity will be change in

$Z.dir^n$ and keeping constant in

$x-y \ dir^n$ .

$\therefore V_{net} = \sqrt{V_0^2+V_0^2 + \left(\left( \dfrac{eE_0}{m}\right)\right)^2}$

$\Rightarrow P = \dfrac{h}{\lambda}$

$\Rightarrow mV_{net} = \dfrac{h}{\lambda}$

$\Rightarrow \lambda = \dfrac{h}{mV_{net}} = \dfrac{h}{m\sqrt{2V_0^2 + \dfrac{e^2E_0^2}{m^2}t^2}}$

$\left( \because \lambda_0 = \dfrac{h}{m\sqrt{2}V_0}\right)$

$\Rightarrow \lambda = \dfrac{h}{\sqrt{\left(1 + \dfrac{e^2E_0^2}{2m^2V_0^2}t^2\right)}}$

The allowed energy for the particle for a particular value of n is proportional to :

0%
$\mathrm{a}^{-2}$
0%
$\mathrm{a}^{-3/2}$
0%
$\mathrm{a}^{-1}$
0%
$\mathrm{a}^{2}$

Explanation

$a = \cfrac{n \lambda}{2}$
De broglie wavelength

$\lambda = \cfrac{h}{mv}$
Subsitute

$p =\cfrac{nh}{2a}$

$E = \cfrac{1}{2}mv^2= \cfrac{p^2}{2m} =\cfrac{n^2h^2}{8a^2m}$

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30

$mW$ and the speed of light is

$3\times 10^{8}m/s$ . The final momentum of the object is:

0%
$0.3\times 10^{-17}kgms^{-1}$
0%
$1.0\times 10^{-17}kg ms^{-1}$
0%
$3.0\times 10^{-17}kgms^{-1}$
0%
$9.0\times 10^{-17}kgms^{-1}$

Explanation

Sol

$. (B)$

$t=100\times 10^{-9} sec,\ P=30\times 10^{-3}$ Watt,

$C=3\times 10^{8}m/s$

Incident energy

$= Pt$

By debroglie wavelength,

$\lambda=h/p$

and Energy of photon is

$E=hc/\lambda$

From above two equations,

$E=pc$

Momentum

$p=\dfrac{Pt}{C}=\dfrac{30\times 10^{-3}\times 100\times 10^{-9}}{3\times 10^{8}}=1.0\times 10^{-17}$ kg ms

$^{-1}$

So, the answer is option (B).

Emission of photoelectrons will not take place light if two different frequencies, whose photons have energy 1 electron volt and 2.5 electron volt are incident one by one on a metal surface of work function 0.5 electron volt. The ratio of maximum energy of emitted electrons will be ?

0%
1:4
0%
4:1
0%
1:2
0%
2:1

The speed of the particle that can take discrete values is proportional to

0%
$\mathrm{n}^{-3/2}$
0%
$\mathrm{n}^{-1}$
0%
$\mathrm{n}^{1/2}$
0%
$n$

Explanation

As given in the question
Let the no. of loops in the wave be

$n$
Let wavelength be

$\lambda$
Debroglie wavelength

$\lambda = \cfrac{h}{mv}$ ------------(1)
Also

$n \lambda = 2 \pi a$ --------------(2)
Substituting above

$v = \cfrac{nh}{2am}$

$v \propto n$

So, the answer is option (D).

Light of wavelength

$\displaystyle { \lambda }_{ ph }$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is

$\displaystyle \phi$ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is

$\displaystyle { \lambda }_{ e }$ , which of the following statement(s) is(are) true?

0%
$\displaystyle { \lambda }_{ e }$ increases at the same rate as ${ \lambda }_{ ph }$ for $\lambda_{ph} < hc / \phi$
0%
$\displaystyle { \lambda }_{ e }$ is approximately halved, if d is doubled
0%
$\displaystyle { \lambda }_{ e }$ decreases with increase in $\displaystyle \phi$ and $\displaystyle { \lambda }_{ ph }$
0%
For large potential difference $\displaystyle \left( V>>\phi /e \right) { \lambda }_{ e }$ is approximately halved if V is made four time

Explanation

${ \lambda }_{ e }=$ minimum be broglie wavelength.

$\dfrac { hc }{ { \lambda }_{ ph } } =\dfrac { hc }{ { \lambda }_{ e } } +\phi +eV$

$\Rightarrow$

$hc(\dfrac { -1 }{ { \lambda }_{ ph }^{ 2 } } d{ \lambda }_{ ph })=hc(\dfrac { -1 }{ { \lambda }_{ e }^{ 2 } } d{ \lambda }_{ e })$

$\Rightarrow$

$\dfrac { d{ \lambda }_{ ph } }{ d{ \lambda }_{ e } } =\dfrac { { \lambda }_{ ph }^{ 2 } }{ { \lambda }_{ e }^{ 2 } }$

This denies option A, B and C.

$V>>\dfrac { \phi }{ e }$

Energy of electron,

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

$\lambda_e =\dfrac { h }{ p } =\dfrac { h }{ \sqrt { 2mE } }$

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

This supports option D.

Electrons of mass m with de-Broglie wavelength

$\lambda$ fall on the target in an X-rays tube. The cutoff wavelength

$(\lambda_0)$ of the emitted X-rays is

0%
$\lambda_0=\lambda$
0%
$\lambda_0=\dfrac{2mc\lambda^2}{h}$
0%
$\lambda_0=\dfrac{2h}{mc}$
0%
$\lambda_0=\dfrac{2m^2c^2\lambda^3}{h^2}$

Explanation

Using de-Broglie equation

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

$p =\sqrt{2mE}$

$\implies$

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

Energy of the X-ray emitted

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

$\therefore$

$\lambda = \dfrac{h}{\sqrt{2m \times \dfrac{hc}{\lambda_o}}}$

$\implies \lambda_o = \dfrac{2mc\lambda^2}{h}$

A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (

$C=$ Velocity of light)

0%
$\frac {2E}{C^2}$
0%
$\frac {E}{C^2}$
0%
$\frac {E}{C}$
0%
$\frac {2E}{C}$

Explanation

Hint: In the relativistic case the momentum is defined as the ratio of the energy and the speed of light.

Step1: Finding the transferred momentum

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

When light gets reflected perfectly, change in momentum is,

$\Delta \mathrm{P}=2 \dfrac{\mathrm{E}}{\mathrm{C}}$

$\therefore$ Momentum transferred to the surface is $\dfrac{\mathrm{2E}}{\mathrm{C}}$ (D)

Light of wavelength

$500\ nm$ is incident on a metal with a work function

$2.28\ eV$ . The de Borglie wavelength of the emitted electron is:

0%
$\leq 2.8 \times 10^{-12}m$
0%
$< 2.8 \times 10^{-10}m$
0%
$< 2.8 \times 10^{-9}m$
0%
$\geq 2.8 \times 10^{-9}m$

Explanation

From photoelectric effect,

$h\nu=W_0+eV$

$\dfrac{hc}{\lambda}=W_0+eV$

$\displaystyle \frac{(6.6\times 10^{-34})\times (3\times 10^8)}{500\times 10^{-9}}=2.28\times 1.6\times 10^{-19}+eV$

$eV=0.31 \times 10^{-19}$ or

$V=0.2$

Here V should be

$V\leq 0.2$

The de Broglie wavelength of electron ,

$\lambda=\dfrac{12.27\times 10^{-10}}{V}$

so,

$\lambda\geq 2.8\times 10^{-9} m$

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is :

0%
25
0%
75
0%
60
0%
50

Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?

Explanation

HINT: ACCORDING TO DE BROGLIE THE WAVELENGTH OF AN OBJECT IS ASSOCIATED WITH THE MOMENTUM AND MASS OF THE BODY

STEP 1: Write the formula of momentum.

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

As $\lambda$ increases, $p$ decreases.

Step 2: Find the correct graph.

Option(A) is wrong since the slope is a straight line but according to the above relation, the graph can not be a straight line.

Differentiating the momentum.

$\dfrac{d p}{d x}=-\dfrac{h}{\lambda^{2}}$

as the value is negative therefore the graph should be decreasing.

Therefore, option (D) is the correct answer.

An electron is accelerated from rest through a potential difference of

$V$ volt. If the de Broglie wavelength of the electron is

$1.227 \times 10^{-2} nm$ , the potential difference is :

0%
$10^2V$
0%
$10^3V$
0%
$10^4V$
0%
$10V$

Explanation

The de-broglie wavelength of an electron is given as:

$\lambda=\dfrac{1.227}{\sqrt{V}}\ nm$

Substitute the wavelength in the above expression:

$V=\left(\dfrac{1.227}{1.227\times10^{-2}}\right)^2$

$V=10^4\ V$

An electron of mass

$m$ and a photon have same energy

$E$ . The ratio of de-Broglie wavelength associated with them is:

0%
$\dfrac { 1 }{ c } { \left( \dfrac { E }{ 2m } \right) }^{ \dfrac { 1 }{ 2 } }$
0%
${ \left( \dfrac { E }{ 2m } \right) }^{ \dfrac { 1 }{ 2 } }$
0%
$c{ \left( 2mE \right) }^{ \dfrac { 1 }{ 2 } }$
0%
$\dfrac { 1 }{ c } { \left( \dfrac { 2m }{ E } \right) }^{ \dfrac { 1 }{ 2 } }$

( $c$ being velocity of light)

Explanation

For an electron,

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

For photon,

$E=pc\Rightarrow \lambda_{Photon}=\dfrac{hc}{E}$

$\Rightarrow \dfrac{\lambda_e}{\lambda_{Photon}}=\dfrac{h}{\sqrt{2mE}}\times \dfrac{E}{hc}=\left (\dfrac{E}{2m} \right )^{1/2}\dfrac{1}{c}$

The wavelength

$\lambda _{e}$ of an electron and

$\lambda _{p}$ of a photon of same energy E are related by :

0%
$\lambda _{p}\propto \lambda _{e}^{2}$
0%
$\lambda _{p}\propto \lambda _{e}$
0%
$\lambda _{p}\propto \sqrt{\lambda _{e}}$
0%
$\displaystyle \lambda _{p}\propto \dfrac{1}{\sqrt{\lambda _{e}}}$

Explanation

Wavelength of electron,

$\displaystyle \lambda _{e}=\dfrac{h}{\sqrt{2mE}}$

Wavelength of photon,

$\displaystyle \lambda _{p}=\dfrac{hc}{E}$

$\displaystyle \lambda _{e}^{2}=\dfrac{h^{2}}{2mE}$

$\displaystyle \lambda _{e}^{2}=\dfrac{h^{2}}{2m\dfrac{hc}{\lambda _{p}}}\\ \Rightarrow \lambda _{e}^{2}\propto \lambda _{p}$

If the momentum of an electron is changed by

$P$ , then the de-Broglie wavelength associated with it changes by

$0.5%$ . The initial momentum of electron will be :

0%
$400P$
0%
$\cfrac{P}{200}$
0%
$100P$
0%
$200P$

Explanation

Initial momentum,

$P_i=\cfrac{h}{\lambda}$
or

$P_i-P=\dfrac{h}{\lambda+\dfrac{0.5}{100}\lambda}$
or

$P_i-P=\dfrac{P_i}{\dfrac{1005}{1000}}$
or

$P_i=\dfrac{1005P}{5}\approx 200P$

If the kinetic energy of a particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is:

0%
25%
0%
75%
0%
60%
0%
50%

Explanation

Let the initial de Broglie wavelength be

$\lambda_0=\dfrac{h}{mv_0}$

The kinetic energy is increased 16 times.

Thus

$KE'=16KE_0$

$\implies \dfrac{1}{2}mv^2=16\times \dfrac{1}{2}mv_0^2$

$\implies v=4v_0$

Thus the new de Broglie wavelength=

$\lambda=\dfrac{h}{mv}=\dfrac{h}{m(4v_0)}=\dfrac{\lambda_0}{4}$

Thus the percentage change in wavelength:

$=\dfrac{\lambda_0-\lambda}{\lambda_0}\times 100\\=75$ %

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

de Broglie wavelength

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

$p = mv$

$\implies$

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

Thus implies

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

$(\because$

$v$ and

$h$ are constant

$)$

Hence Reason statement is correct.

Mass of microscopic and sub-microscopic particle is very small. Thus according to de broglie equation, these particles tend to behave like a wave with certain finite wavelength which suggests the wave-particle dual nature of such particles like electrons.Thus de Broglie equation has significance for microscopic and sub -microscopic particles.

Thus both Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

De Broglie wavelength

$\lambda$ associated with neutrons is related with absolute temperature T as:

0%
$\lambda \propto T$
0%
$\lambda \propto \frac{1}{T}$
0%
$\lambda \propto \frac{1}{\sqrt{T}}$
0%
$\lambda \propto T^2$

Explanation

De broglie wavelength

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

where

$v$ is the velocity of the particle.

Let

$E$ be the kinetic energy of the neutrons.

Thus

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

$\implies v=\sqrt{\dfrac{2E}{m}}$

Therefore

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

Since energy of neutrons

$E\propto T$

$\lambda\propto \dfrac{1}{\sqrt{T}}$

If we assume kinetic energy of a proton is equal to energy of the photon, the ratio of de Broglie wave length of proton to photon is proportional to :

0%
$\displaystyle E$
0%
$\displaystyle { E }^{ { -1 }/{ 2 } }$
0%
$\displaystyle { E }^{ { 1 }/{ 2 } }$
0%
$\displaystyle { E }^{ { 3 }/{ 2 } }$

If velocity of a particle is three times of that of electron and ratio of de Broglie wavelength of particle to that of electron is

$1.814 \times 10^{-4}$ . The particle will be :

0%
Neutron
0%
Deutron
0%
Alpha
0%
Tritium

Explanation

de broglie wavelength,

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

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

$\Rightarrow \dfrac {\lambda_{p}}{\lambda_{e}} = \dfrac {m_{e}v_{e}}{m_{p}v_{p}}$

$\Rightarrow m_{p} = \dfrac {m_{e}v_{e}}{v_{p}}\times \dfrac {\lambda_{e}}{\lambda_{p}}$

Here,

$m_{e} = 9.1\times 10^{-31} kg, v_{p} = 3v_{e}$ and

$\dfrac {\lambda_{p}}{\lambda_{e}} = 1.814 \times 10^{-4}$

$\Rightarrow m_{p} = \dfrac {9.1\times 10^{-31}}{1.814\times 10^{-4} \times 3} = 1.672\times 10^{-27} kg$

Thus, the particle is neutron.

Monochromatic light of wavelength

$3000\overset {\circ}{A}$ is incident on a surface area

$4\ cm^{2}$ . If intensity of light is

$150\ mW/m^{2}$ , then rate at which photons strike the target is

0%
$3\times 10^{10}/s$
0%
$9\times 10^{13}/s$
0%
$7\times 10^{15}/s$
0%
$6\times 10^{19}/s$

Explanation

$\dfrac {n}{t} = \dfrac {lA\lambda}{hc}$

$= \dfrac {150\times 10^{-3}\times 4\times 10^{-3}\times 3\times 10^{-7}}{6.6\times 10^{-34}\times 3\times 10^{8}}$

$= 9\times 10^{13}/ s$ .

The de Broglie wavelength of a neutron when its kinetic energy is

$K$ , is

$\lambda$ . What will be its wavelength when its kinetic energy is

$4K$ ?

0%
$\dfrac{\lambda}{4}$
0%
$\dfrac{\lambda}{2}$
0%
$2\lambda$
0%
$4\lambda$

Explanation

Kinetic energy of neutron=

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

$\implies v=\sqrt{\dfrac{2K}{m}}$

De-broglie wavelength of neturon=

$\dfrac{h}{p}$

$=\dfrac{h}{mv}=\dfrac{h}{\sqrt{2mK}}\propto \dfrac{1}{\sqrt{K}}$

Thus when kinetic energy becomes

$4K$ , the de-broglie wavelength becomes

$\dfrac{\lambda}{2}$ .

The photoelectric threshold energy of certain metal is 3 eV. If light of wavelength 3000

$A^{o}$ is incident on the metal, then :

0%
Electrons will be emitted
0%
Positrons will be emitted
0%
Protons will be emitted
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Electrons will not be emitted

Explanation

Energy of incident rays

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

$=\dfrac{1240}{300} \ \ \ \ (hc=1240 eV-nm)$

$=4.13eV$
So, energy of incident rays is more than the threshold energy (3ev) so, electrons emission takes place.
So, the answer is option (A).

An electron beam after collision with the target produces X-rays of wavelength

$4 A^0$ . The velocity of the electron beam is

0%
$3.31\times 10^{7}m/s$
0%
$6.31\times 10^{7}m/s$
0%
$8.31\times 10^{7}m/s$
0%
$9.31\times 10^{7}m/s$

Explanation

Energy of x-ray beam

$= \frac{hc}{ \lambda} = 4.95*10^{-16} J =\frac{1}{2} mv^2$ and it gives

$v = 6.31*10^7$ m/s.

So, the answer is option (B).

The momentum of a proton is p. The corresponding wavelength is

0%
h/p
0%
hp
0%
p/h
0%
$\sqrt{hp}$

Explanation

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

$=\dfrac{h}{p}$

Moving with the same velocity, one of the following has the longest de Broglie wavelength

0%
$\beta -$ particle
0%
$\alpha -$ particle
0%
proton
0%
neutron

Explanation

De-broglie wavelength

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

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

$\beta$ particle has least mass.

So $\beta$ particle has the longest wavelength.

A photon of frequency

${\text{x}}$ has an energy

0%
$\dfrac{h}{x^{2}}$
0%
$\dfrac{x}{h}$
0%
$hx$
0%
$hx^{2}$

Explanation

Energy of photon is

$hv$ , where

$v$ is its frequency.

The wavelength of matter waves does not depend on

0%
Momentum
0%
Velocity
0%
Mass
0%
Charge

Explanation

Wavelength of matter,

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

De-Broglie wavelength depends on

0%
Mass of the particle
0%
Size of the particle
0%
Material of the particle
0%
Shape of the particle

Explanation

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

$\lambda$ depends on mass of the particle.

If the energy and momentum of a photon are E and P respectively, then the velocity of photon will be:

(one or more than one correct)

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$E/P$
0%
$(E/P)^{2}$
0%
$EP$
0%
$3\times 10^{8}m/s$

Explanation

Answers are (A) and (D).

Energy of Photon E

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

momentum of Photon

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

$\therefore \dfrac{E}{P}=\dfrac { \dfrac { hc }{ \lambda } }{ \dfrac { h }{ \lambda } } =c=$ Velocity of Photon =

$3\times 10^{8}m/s$

The effective mass of photon in microwave region, visible region and x -ray region is in the following order:

0%
X-rays > Visible > Microwave
0%
Microwave > X-rays > Visible
0%
X-rays > Microwave > Visible
0%
Microwave > Visible> X-ray

Explanation

Relation between wavelength and mass is given as

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

Since h and v are constant, higher the wavelength less the mass wavelength of x ray is least followed by visible ray and then microwave.

So effective mass of photon follows

$X-rays>Visible>Microwave$

So, the answer is option (D).

Matter waves are:

0%
Electromagnetic waves
0%
Mechanical waves
0%
Either mechanical or electromagnetic waves
0%
Neither mechanical nor electromagnetic waves

Explanation

Correct Answer: option (D)

Matter waves are waves associated with every moving particle and therefore are neither mechanical nor electromagnetic waves.

These waves have

$\lambda = h\times p$

Where $\lambda$ is the wavelength, $h$ is plank's constant, and $p$ is the particle’s moment.

Therefore, matter waves are neither mechanical nor electromagnetic waves.

The frequency of a photon associated with an energy of 3.31 eV is (given h

$=6.625\times 10^{-34}$ Js)

0%
$0.8\times 10^{15}$
0%
$1.6\times 10^{15}$
0%
$3.2\times 10^{15}$
0%
$8.0\times 10^{15}$

Explanation

energy

$= h\nu$

$3.31eV = 6.625*10^{-34}\nu$

$\nu = \dfrac{3.31*1.6*10^{-19}}{6.625*10^{-34}}$ (

$\because e =1.6*10^{-19}C$ )

$= 0.8*10^{15}Hz$

A proton when accelerated through a potential difference of V volt has a wavelength

$\lambda$ associated with it. An

$\alpha$ - particle in order to have the same wavelength

$\lambda$ must be accelerated through a p.d. of

0%
V/8 volt
0%
V/4 volt
0%
V volt
0%
2V volt

Explanation

For proton,

$\lambda_p =\dfrac{0.286\times 10^{-10}}{\sqrt{\nu }}\ m$
where V is p.d
For

$\alpha -particle,\ \lambda_\alpha =\dfrac{0.101\times 10^{-10}}{\sqrt{\nu }}m$

so, keeping same

$\alpha$ for proton and

$\alpha -partside$

$1= \dfrac{\lambda _p}{\lambda _{\alpha }} = \dfrac{0.286 \times \sqrt{V_{\alpha }}}{0.101 \times \sqrt{V_{P}}}$

$0.101 \times \sqrt{V_{P}} = 0.286 \times \sqrt{V_{\alpha }}$

$(0.101)^{2} \times V_{P} = (0.286)^{2}V _{\alpha}$

$\dfrac{(0.101)^{2}\nu }{(0.286)^{2}} = V _{\alpha }$

$(\because V_{P}=V)$

$\Rightarrow V_{\alpha } = \dfrac{V}{8} volt$

So, the answer is option (A).

The wavelength associated with an electron having kinetic energy is given by the expression:

0%
$h/\sqrt{2mE}$
0%
$2h/mE$
0%
$2 mhE$
0%
$\dfrac{2\sqrt{2mE}}{h}$

Explanation

$\dfrac{1}{2}mv^{2}=E\ (given)$

$mv=\sqrt{2Em}$

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

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

So, the answer is option (A).

De Broglie wavelength ‘

$\lambda$ ’ is proportional to

0%
$\dfrac{1}{\sqrt{E}}$ for photons and $\dfrac{1}{E}$ for particles
0%
$\dfrac{1}{E}$ for photons and $\dfrac{1}{\sqrt{E}}$ for particles
0%
$\dfrac{1}{E}$ for both photons and particles in motion
0%
$\dfrac{1}{\sqrt{E}}$ for both photons and particles

Explanation

For photon

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

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

$\therefore \ \lambda \ \alpha \dfrac{1}{E}$

and for partides

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

$mv = \sqrt{2Em}$

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

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

$\lambda \alpha \dfrac{1}{\sqrt{E}}$

So, the answer is option (B).

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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