MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 15

An electron and a proton have the same De Broglie wavelength. Then the kinetic energy of the electron is :

0%
Zero
0%
Infinity
0%
Equal kinetic energy of proton
0%
Greater than the kinetic energy of proton

Explanation

We know that,

$E=hv=\dfrac { { h }_{ c } }{ \lambda }$ is the fundamental relation.

$E=cp$

$\Rightarrow cp=\dfrac { { h }_{ c } }{ \lambda }$ or

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

Thus, it is evident that same de-Broglie wavelength implies, Total energies of the particles are equal.

Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is

${ V }_{ 0 }$ and the maximum kinetic energy of the photoelectrons is

${ K }_{ max }$ . When the ultraviolet light is replaced by X-rays, both

${ V }_{ 0 }$ and

${ K }_{ max }$ increase.
Statement-2 : photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

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Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1.
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Statement-1 is true, Statement-2 is true: Statement-2 is not the correct explanation of Statement-1
0%
Statement-1 is false, Statement-2 is true.
0%
Statement-1 is true, Statement-2 is false.

Explanation

Since the frequency of ultraviolet light is less than the frequency of X-rays, the

energy of each incident photon will be more for X-rays

KEphotoelectron=hv–ϕ

Stopping potential is to stop the fastest photoelectron

V∘=

$\dfrac{hv}{e}–\dfrac{ϕ}{e}$

and V∘ both increases. So, KEmax

But KE ranges from zero to KEmax because of loss of energy

due to subsequent collisions before getting ejected and not due to

range of frequencies in the incident light.

hence option D is the correct answer

$\left( {{\text{radius}}\;2\;\mathop {\text{A}}\limits^0 } \right)$ in the cathode to accumulate sufficient energy to eject a photoelectron?

0%
355 seconds
0%
176 seconds
0%
704 seconds
0%
No time lag

A beam of light has three wavelengths

$4144\mathring{A}$ ,

$4972\mathring { A }$ and

$6216\mathring { A }$ with a total intensity of

$3.6\times { 10 }^{ -3 }\quad { Wm }^{ -2 }$ equally distributed among the two wavelengths. The beam fall normally on area of

$1\ { cm }^{ 2 }$ of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in 2s is approximately

0%
$6\times { 10 }^{ 11 }$
0%
$9\times { 10 }^{ 11 }$
0%
$11\times { 10 }^{ 11 }$
0%
$15\times { 10 }^{ 11 }$

Explanation

$E_1 = \dfrac{12400}{4144}$ = 2.99 eV.

$E_2 = \dfrac{12400}{4972}$ = 2.49 eV.

$E_3 = \dfrac{12400}{6216}$ = 1.99 eV.

Only the first two wavelengths are capable of ejecting photoelectrons.

Energy incident per second

$\dfrac{3.6}{3} \times{10^{-3}} \times{10^{-4}} = 1.2 \times{10^{-7}}$ J /s

$n_1 = \dfrac{1.2 \times{10^-7}}{2.99 \times {1.6 \times{10^{-19}}}} = 2.5 \times {10^{11}}$

$n_2 = \dfrac{1.2 \times {10^{-7}}}{2.99\times{1.6\times{10^{-19}}}} = 3 \times {10^{11}}$

Total number of photons = 2

$(n_1 + n_2)$

= 3.01

$\times {1011} + 2.51 \times{011} = 5.52 \times{10^{11}}$

Total number of photo electrons ejected in two seconds =

$11 \times{10^{11}}$

An electron passing through a potential difference of 4.9 V collides with a mercury atom and transfer it to the first excited state. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state?

0%
2050 $A^0$
0%
2240 $A^0$
0%
2540 $A^0$
0%
2525 $A^0$

Explanation

Given,

$h=6.6\times 10^{-34}$

$c=3\times 10^8$

$e=1.6\times 10^-19$

$v=4.9$

we know that

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

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

$\therefore \lambda=\dfrac{6.6\times 10^{-34}\times 3\times 10^8}{1.6\times 10^-19 \times 4.9}$

$\lambda=2525^{o}A$

The following diagram indicates the energy levels of a certain atom when the system moves from

$4E$ level to

$E$ . A photon of wavelength.

$\lambda_1$ is emitted. The wavelength of photon produced during its transition from

$\displaystyle \dfrac{7}{3}E$ level to

$E$ is

$\lambda_2$ . The ratio

$\displaystyle \dfrac{\lambda_1}{\lambda_2}$ will be

0%
$\displaystyle \dfrac{9}{4}$
0%
$\displaystyle \dfrac{4}{9}$
0%
$\displaystyle \dfrac{3}{2}$
0%
$\displaystyle \dfrac{7}{3}$

Explanation

$4E-E=\dfrac{hc}{{\lambda}_{1}}$

$\Rightarrow 3E=\dfrac{hc}{{\lambda}_{1}}$ ..........

$(1)$

$\dfrac{7E}{3}-E=\dfrac{hc}{{\lambda}_{2}}$

$\Rightarrow \dfrac{7E-3E}{3}=\dfrac{hc}{{\lambda}_{2}}$

$\Rightarrow \dfrac{4E}{3}=\dfrac{hc}{{\lambda}_{2}}$ ..........

$(2)$

Dividing eqns

$(1)$ by

$(2)$ we get

$\dfrac{3E}{\dfrac{4E}{3}}=\dfrac{\dfrac{hc}{{\lambda}_{1}}}{\dfrac{hc}{{\lambda}_{2}}}$

$\Rightarrow \dfrac{{\lambda}_{1}}{{\lambda}_{2}}=\dfrac{9}{4}$

Light of wavelength

$1800\overset { \circ }{ A }$ ejects photoelectrons from plate of a metal whose work function is

$2eV$ . Electron ejected from the plate with maximum kinetic energy

0%
$4.88eV$
0%
$6.72eV$
0%
$1.43eV$
0%
$2.54eV$

Explanation

Energy of incident light =

$\dfrac{hc}{\lambda}$

${\lambda}$ = 180

${\buildrel_{\circ}\over{\mathrm{A}}}$

we know that

hv =

$\phi_0 +\dfrac{1}{2}\times{m}{v^2}$

$\dfrac{hc}{\lambda} = \phi_1 +\dfrac{1}{2}\times{m}{v^2}$

$K.E_{max} = \dfrac{hc}{\lambda} - \phi_0$

$0.011\times{10^{-16}} - 3.2\times{10^{-19}} J$

$7.3\times{10^{-19}} J$

= 1.43 ev

Photons of wavelength 6620 A are incident normally on a perfectly reflecting screen. Calculate the number of photons per second falling on the screen as total power of photons such that the exerted force is 1 N:

0%
$5\times { 10 }^{ 26 }$
0%
$5\times { 10 }^{ 25 }$
0%
$1.5\times { 10 }^{ 8 }$
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None of these

The de Broglie wavelengths associated with a proton and an

$\alpha$ - particle will be:

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

The energy of a $k$ electron in tungsten is -20 KeV and that of an $l$ Electron is -2 KeV. Find the wavelength of X rays emitted when there is electron jump from L to K shell.

0%
$0.3443\mathop {\text{A}}\limits^0$
0%
$0.6887\mathop {\text{A}}\limits^0$
0%
$1.3982\mathop {\text{A}}\limits^0$
0%
$2.87\mathop {\text{A}}\limits^0$

When a potential drop of

$V$ volt is applied on a stationary neutron, then de broglie wavelength:-

0%
$0$
0%
$\infty$
0%
$\dfrac{12.27}{\sqrt V} \mathring{A}$
0%
$\dfrac{0.101}{\sqrt V} \mathring{A}$

If given particles are moving with same velocity, then maximum de-Broglie wavelength for

0%
Proton
0%
$\alpha$ -particle
0%
Neutron
0%
$\beta$ -particle

$X$ - rays tube operate at

$10 KV$ . The ratio of X- rays wavelength to that of De - Broglie is

0%
$1:10$
0%
$10:1$
0%
$1 : 100$
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$100: 1$

The wavelength of very fast moving electron

$\left( {v \approx c} \right)$ is

$:$

0%
$\left( {v \approx c} \right)$
0%
$\lambda = \frac{h}{{\sqrt {2mE} }}$
0%
${\lambda ^2} = \frac{{{h^2}}}{{\sqrt {2mE} }}$
0%
$\lambda = \frac{{h\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}{{{m_0}v}}$

If accelerating voltage of X-ray tube is

$13 kv$ find minimum wavelength of X-ray

$I=\dfrac{12400}{13k}$

0%
$1\mathring{A}$
0%
$0.82\mathring{A}$
0%
$0.95\mathring{A}$
0%
$1.72\mathring{A}$

A photon was absorbed by a hydrogen atom in its ground state, and the electron was prompted to the fifth orbit. When the excited atom returned
to its ground state, visible and other quanta were emitted. In this process, how many maximum spectral lines could be obtained-

0%
1
0%
2
0%
5
0%
10

Explanation

The emission of visible light invoice transmission to second orbit as it invoice Balmer series (from fifth to second orbital). Thus, emitted atom is second orbital, transmission must be from

$2\rightarrow 1$ for an atom to return to its ground state.

What is the energy (approximate) of a photon emitted when an electron in a doubly charged lithium ion

${ LI }^{ ++ }$ (with nuclear charge 3e) undergoes transition between n=3 and n=1 states (n being the principal quantum number)

0%
108.8 eV
0%
13.6 eV
0%
10.9 eV
0%
122.4 eV

The momentum of the photon of wavelength

$5000\overset {\circ}{A}$ will be

0%
$1.3\times 10^{-27} kg-m/s$
0%
$1.3\times 10^{-28} kg-m/s$
0%
$4\times 10^{29} kg-m/s$
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$4\times 10^{-18} kg-m/s$

$n_R$ and

$n_g$ given the number of photons in red and green beams of light. If the two beams have the same energy, then,

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$n_R < n_g$
0%
$n_R > n_g$
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$n_R = n_g$
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all could be true

The increase in the energy of an electron for changing its de-Broglie wavelength from

$1$

$\overset { \circ }{ A }$ to

$0.5$

$\overset { \circ }{ A }$ will be -

0%
1.5 keV
0%
0.45 keV
0%
0.25 keV
0%
100 eV

${\lambda }_{ e }$ ,

${ \lambda }_{ p}$ and

${ \lambda }_{a }$ are the de broglie wavelengths of electrons , proton and

$\alpha$ particles. If all are accelerated by same potential , then.

0%
${ \lambda }_{ e }<{ \lambda }_{ p } <{ \lambda }_{ a }$
0%
${ \lambda }_{ e }<{ \lambda }_{ p } >{ \lambda }_{ a }$
0%
${ \lambda }_{ e }>{ \lambda }_{ p } <{ \lambda }_{ a }$
0%
${ \lambda }_{ e }>{ \lambda }_{ p } >{ \lambda }_{ a }$

Momentum of

$\gamma$ -ray photon of energy 3 keV in km/s will be

0%
$1.6 \times 10^{-19}kgms^-1$
0%
$1.6 \times 10^{21}kgms^-1$
0%
$1.6 \times 10^{-24}kgms^-1$
0%
$1.6 \times 10^{-27}kgms^-1$

Explanation

$\begin{array}{l} Momentum\, \, of\, \, photon=\dfrac { h }{ \lambda } \\ Energy\, \, of\, \, photon=\dfrac { { hc } }{ \lambda } \\ 3KeV =\dfrac { { hc } }{ \lambda } \\ \frac { h }{ \lambda } =3\times 1.6\times { 10^{ -19 } }\times { 10^{ 3 } } \\ \dfrac { h }{ \lambda } =3\times 1.6\times { 10^{ -19 } }\times { 10^{ 3 } }\times { 10^{ -3 } } \\ =1.6\times { 10^{ -19 } }. \end{array}$

Option A.

A photon is incident having frequency

$1\times { 10 }^{ 14 }{ sec }^{ -1 }$ . Threshold frequency of metal

$5\times { 10 }^{ 13 }{ sec }^{ -1 }$ . Find the kinetic energy of the ejection electron :-

0%
$3.3\times { 10 }^{ -21 }$ J
0%
$6.6\times { 10 }^{ -21 }$ J
0%
$3.3\times { 10 }^{ -20 }$ J
0%
$6.6\times { 10 }^{ -20 }$ J

A photon of red light having wavelength

$660\ nm$ has energy equal to

$\left( h=6.6\times { 10 }^{ -34 }Js \right)$

0%
$1.0\times { 10 }^{ -19 }J$
0%
$3.0\times { 10 }^{ -18 }J$
0%
$1.0\times { 10 }^{ -9 }J$
0%
$3.0\times { 10 }^{ -19 }J$

An electron is confined to tube of length L, the electron's potential energy in one half of the tube is zero, while the potential energy in the other half is 10eV. if the electrons has a total energy E= 15 eV, then the ratio of the de brogile wavelength of the electron in the 10 eV region of the tube to that inthe other half is :-

0%
$\frac {1}{\sqrt 3}$
0%
$\sqrt 3$
0%
$3$
0%
$\frac {1}{3}$

The following graph shows the variation of photo current for a photosensitive metal.

0%
Identify the variable $X$ on the horizontal axis.
0%
What does the point $A$ on the horizontal axis represent?
0%
Draw this graph for three different values of frequencies of incident radiation $v_{1},v_{2}$ and $v_{3}(v_{1}>v_{2}>v_{3})$ for the same intensity.
0%
Draw this graph for three different values of intensities of incident $I_{1},I_{2}$ and $I_{3}(I_{1}>I_{2}>I_{3})$

In which of the following transition will the wavelength be minimum?

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n=2 to n= 1
0%
n= 3 to n=2
0%
n= 4 to n=3
0%
n=5 to n=4

The wavelength

$\lambda _e$ , of an electron and

$\lambda _p$ of a photon of same energy

$E$ are related by

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

The velocity of an electron in the ground state of hydrogen atom is

$2.2\times 10^{6}m/s$ . The De-Broglie wavelength associated with a muon in the ground state of a muonic hydrogen will be -

$(m\mu =207 m_{e} , h = 6.62 \times 10^{-34}Js)$

0%
$1.6\mathring { A }$
0%
$0.16\mathring { A }$
0%
$0.016\mathring { A }$
0%
$0.0016\mathring { A }$

An electron, proton and alpha particle have same
kinetic energy. The corresponding de-Broglie wavelength would have the
following relationship:

0%
$\lambda _ { \mathrm { e } } >\lambda _ { \mathrm { p } } > \lambda _ { \alpha }$
0%
$\lambda _ { \mathrm { p } } < \lambda _ { \mathrm { e } } > \lambda _ { \alpha }$
0%
$\lambda _ { \alpha } < \lambda _ { \mathrm { e } } > \lambda _ { \mathrm { p } }$
0%
$\lambda _ { \alpha } ^ { r } < \lambda _ { p } > \lambda _ { \mathrm { e } }$

During an

$'\alpha '$ particle decay from a nuclei at rest of atomic mass 284, the wavelength of the wave associated with the

$'\alpha '$ particle is

$'\lambda '$ . Then the wavelength of the wave associated with the daughter nuclei must be

0%
$\lambda$
0%
$\dfrac { \lambda }{ 2 }$
0%
$2\lambda$
0%
$280\lambda$

The frequency and wavelength of radiation which has energy of 5 eV is

0%
$1.2 \times 10^{15}\ Hz, 2848\ \overset { \circ }{ A }$
0%
$1.2 \times 10^{13}\ Hz, 2488\ \overset { \circ }{ A }$
0%
$1.2 \times 10^{15}\ Hz, 2488\ \overset { \circ }{ A }$
0%
$2.1 \times 10^{15}\ Hz, 2488\ \overset { \circ }{ A }$

A perfectly reflecting mirror has an area of

$2\, cm^2$ . Light energy is allowed to fall on it for

$2\,h$ at the rate of

$30\, W\, cm^{-2}$ . The force that acts on the mirror is

0%
$2\times 10^{-7}N$
0%
$4\times 10^{-7}N$
0%
$1.5\times 10^{-8}N$
0%
$4.5\times 10^{-8}N$

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 a-particle will be:

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

A 5 watt source emits monochromatic light of wavelength

$5000 \AA$ . When placed

$0.5 m$ away it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of

$1.0 m$ , the number of photoelectrons liberated will be reduced by a factor of

0%
$8$
0%
$16$
0%
$2$
0%
$1/4$

A proton moves on a circular path of radius

$6.6\times { 10 }^{ -3 }$ m in a perpendicular magnetic field of

$0.625$ tesla. The De broglie wavlength associated with the proton will be:

0%
$1\mathring { A }$
0%
$0.1\mathring { A }$
0%
$0.01\mathring { A }$
0%
$0.001\mathring { A }$

The speed of a proton is

$\dfrac { c }{ 20 } .$ The wavelength associated with it will be(

$h= 6.6 \times 10^{-34} Js$ ) :

0%
$2.64\times { 10 }^{ -24 }mm$
0%
$2.64\times { 10 }^{ -24 }nm$
0%
$2.64\times { 10 }^{ -24 }\mathring { A }$
0%
$2.64\times { 10 }^{ -14 }m$

The ratio of deBroglie wavelengths of a proton and an alpha particle of same energy is

0%
1
0%
2
0%
4
0%
0.25

Explanation

Correct answer: Option (B).

Hint: Apply the formula of de-broglie wavelength in relation to kinetic energy

Step 1: Find out the relation between de -broglie wavelength and kinetic energy.

De – Broglie wavelength is the wavelength that is associated with an object in relation to its momentum and mass.

The de -broglie wavelength can be written as:

$\lambda=\dfrac{h}{p}$ (where, $\lambda$ is the de – Broglie wavelength, $h$ is Planck’s constant and $p$ is the momentum) ….(i)

Now, we know that,

$K.E=\dfrac{p^2}{2m}$ (where, $K.E$ is the kinetic energy)

$\Rightarrow p^2=2mK.E$

$\Rightarrow p=\sqrt{2mK.E}$

Putting the value of $p$ in equation (i), we get,

$\lambda=\dfrac{h}{\sqrt{2mK.E}}$ ….(ii)

Step 2: Calculate the ratio of de-broglie wavelength of a proton and an alpha particle of same energy.

According to the given problem, both the proton and the alpha particle has the same energy.

A proton is designated as $H^1_1$

An alpha particle is designated as $He^4_2$

So, the mass of an alpha particle is four times the mass of a proton.

Let the mass of proton be $m$

Mass of alpha particle will be $4m$

Since, the energy is same and $h$ is constant, equation (ii) can be written as:

$\dfrac{\lambda_{proton}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha}}{m_{proton}}}$

$\Rightarrow\dfrac{\lambda_{proton}}{\lambda_{\alpha}}=\sqrt{\dfrac{4m}{m}}$

$\Rightarrow\dfrac{\lambda_{proton}}{\lambda_{\alpha}}=2$

Hence, the ratio is $2$ .

The correct answer is option (B).

When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x=0 and x=a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as E =

$\frac{\rho^2}{2m}$ . Thus, the energy of the particle can be denoted by a quantum number n taking values 1, 2, 3, . (n = 1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h =

$6.6 \times 10^{-34} \ Js$ and

$e =1.6 \times 10^{-19}$ C.
The speed of the particle that can take discrete values is proportional to

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

Photon having wavelength 310 nm is used to break the bond of

${ A }_{ 2 }\\ \\ \\$ molecule having bond energy 288 kJ

${ mol }^{ -1 }$ then % energy of photon converted to the K,E is [hc=12400 eV, 1 ev=96kJ/mol]

0%
25
0%
50
0%
75
0%
80

The wavelength of de-Broglie waves associated with neutrons at room temperature T is (E=kT)

0%
$\dfrac { 1.82 }{ T } \mathring { A }$
0%
$\dfrac { 1.82 }{ \sqrt { T } } \mathring { A }$
0%
$\dfrac { 30.7 }{ \sqrt { T } } \mathring { A }$
0%
$\dfrac { 30.7 }{ T } \mathring { A }$

A free hydrogen atom after absorbing a photon of wavelength

$\lambda_a$ gets excited from the state

$n = 1$ to the state

$n = 4$ . Immediately after that the electron jumps to n = m state by emitting a photon of wavelength

$\lambda_e$ . Let the change in momentum of atom due to the absorption and the emission are

$\Delta P_a$ and

$\Delta P_e$ respectively. If

$\lambda_a / \lambda_e = 1/5$ , which of the option (s) is/are correct ?
[Use

$hc = 1242 eV$ nm ;

$1 nm = 10^{-9} m$ , h and c are Plank's constant and speed of light, respectively]

0%
the ratio of kinetic energy of the electron in the state n = m to the state $n = 1$ is $1/4$
0%
$m = 2$
0%
$\Delta p_a / \Delta p_e = 1/2$
0%
$\lambda_e = 418 nm$

Explanation

$\dfrac{\lambda_a}{\lambda_e} = \dfrac{E_4 - E_m}{E_4 - E_1} = \dfrac{\dfrac{1}{m^2} - \dfrac{1}{16}}{1 - \dfrac{1}{16}} = \dfrac{1}{5}$

on solving we will get

$m = 2$

$\lambda_e = \dfrac{1242 \times 16}{13.6 \times 3} \approx 487 \, nm$

$\dfrac{K_2}{K_1} = \dfrac{1^2}{2^2} = \dfrac{1}{4}$ as kinetic energy is proportional to

$\dfrac{1}{n^2}$

The de Broglie wavelength is

$\lambda$ , the energy of electron is-

0%
$\frac{h}{\lambda}$
0%
$\frac{h.c}{\lambda}$
0%
$\frac{h^{2}}{2m\lambda^{2}}$
0%
None of these

Explanation

$\begin{array}{l}\lambda=\frac{h}{P}=\frac{h}{m V}\\\therefore V=\frac{h}{m \lambda}\\\therefore\text { Energy }=\frac{1}{2} m v^{2} \\=\frac{1}{2} \times m \times \frac{h^{2}}{m^{2}\lambda^{2}}\\.E=\frac{h^{2}}{2 m\lambda^{2}}\end{array}$

Two particles A and B are moving in same direction have deBroglie wavelengths

${ \lambda }_{ 1 }$ and

${ \lambda }_{ 2 }$ combine to from a particle C. The process conserve momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional)

0%
${ \lambda }_{ 1 }+{ \lambda }_{ 2 }$
0%
$\dfrac { { \lambda }_{ 1 }{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }+{ \lambda }_{ 2 } }$
0%
${ \lambda }_{ 1 }$ and ${ \lambda }_{ 2 }$
0%
$\dfrac { { \lambda }_{ 1 }^{ 2 } }{ { \lambda }_{ 1 }+{ \lambda }_{ 2 } }$

An object in space is moving with a velocity of

$10^{10}cm/s$ towards a star that emits radiations of wavelength

$6\times 10^{-5}cm$ . The wavelength of the radiations recived by the crew in the object will be:

0%
$5.5\times 10^{-5}cm$
0%
$6.5\times 10^{-5}cm$
0%
$4.0\times 10^{-5}cm$
0%
$4.5\times 10^{-5}cm$

De broglie wave length of uncharged particle depends on

0%
Mass of particle
0%
Kinetic energy of particle
0%
Nature of particle
0%
All of the above

The wavelength

$\lambda_e$ an electron and

$\lambda_P$ of a photon of same energy

$E$ are related by:

0%
$\lambda_P\propto\displaystyle\dfrac{1}{1\sqrt{\lambda_e}}$
0%
$\lambda_P\propto\lambda^2_e$
0%
$\lambda_P\propto\lambda_e$
0%
$\lambda_P\propto\sqrt{\lambda_e}$

Only a fraction of the electrical energy supplied to a tungsten light bulb is converted into visible light. If a

$100\ W$ light bulb converts

$20\%$ of the electrical energy into visible light

$(\lambda = 662.6\ nm)$ , then the number of photons emitted by the bulb per second is

0%
$6.67\times 10^{19}$
0%
$2\times 10^{28}$
0%
$6\times 10^{36}$
0%
$6.30\times 10^{19}$

Photo-dissociation of water

$H_2O(1)+hv\to H_2(g)+1/2O_2(g)$ has been suggested as a source of hydrogen. The heat absorbed int his reaction is

$289.5\ kJ/mol$ of water decomposed. The maximum wavelength that would provide the necessary energy assuming that one photon causes the dissociation of one water molecule is

$(1\ eV=96.5\ kJ/mol)$

0%
$6.95\times 10^{-28}m$
0%
$4.19\times 10^{-7}m$
0%
$6.95\times 10^{-31}m$
0%
$1.72\times 10^{-6}m$

Two particles A and B of same mass have their de-Broglie wavelengths in the ratio

$X_A:X_B=K:1$ . Their potential energies

$U_A:U_B=1:K^2$
The ratio of their total energies is

$E_A:E_B$ is

0%
$K^2:1$
0%
$1:K^2$
0%
$K:1$
0%
$1:K$
0%
$1:1$

Explanation

According to question,

$\dfrac{X_A}{X_B}=\dfrac{K}{1}$ .......(i)

$\dfrac{U_A}{U_B}=\dfrac{1}{K^2}$

So,

$U_A=y$ and

$U_B=K^2y$

According to de-Broglie wavelength,

$\lambda=\dfrac{h}{\sqrt{2mK_1}}$ ..........(ii)

Here,

$K_1$ is kinetic energy.

So, from Equations (i) and (ii),

$\Rightarrow \dfrac{\left(\dfrac{h}{\sqrt{2mK_{1A}}}\right)}{\left(\dfrac{k}{\sqrt{2mK_{1B}}}\right)}=\dfrac{K}{1}\Rightarrow \sqrt{\dfrac{K_{1B}}{K_{1A}}}=\dfrac{K}{1}$

$\Rightarrow \dfrac{K_{1B}}{K_{1A}}=K^2$

So,

$K_{1B}=K^2x$ and

$K_{1A}=x$

$\because$ Total energy,

$E=K_1+U$

So,

$E_A=K_{1A}+U_A$

and

$E_B=K_{1B}+U_B$

$\dfrac{E_A}{E_B}=\dfrac{K_{1A}+U_A}{K_{1B}U_{1B}}=\dfrac{x+y}{K^2x+K^2y}$

$\dfrac{E_A}{E_B}=\dfrac{1}{K^2}$ or

$E_A:E_B=1:K^2$

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 15 - 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 15 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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