Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 40

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Weekly Quiz Competition

Question 1
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The speed of an electron having a wavelength of $${ 10 }^{ -10 }m$$ is

A
$$4.24\times { 10 }^{ 6 }m/s\quad $$

B
$$5.25\times { 10 }^{ 6 }m/s$$

C
$$6.26\times { 10 }^{ 6 }m/s\quad $$

D
$$7.25\times { 10 }^{ 6 }m/s\quad $$

Solution

We have
$$\lambda =\cfrac { h }{ p } =\cfrac { h }{ mv } $$
$$\Rightarrow v=\cfrac { h }{ m\lambda } =\cfrac { 6.63\times { 10 }^{ -34 } }{ 9.1\times { 10 }^{ -31 }\times { 10 }^{ -10 } } =7.25\times { 10 }^{ 6 }m/s\quad $$
Question 2
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An electron in an electron microscope with initial velocity $$v_0i$$ enters a region of a stray transverse electric field $$E_0j$$. The time taken for the change in its de Broglie wavelength from the initial value of $$\lambda$$ to $$\dfrac{\lambda}{3}$$ is proportional to.

A
$$E_0$$

B
$$\displaystyle\frac{1}{E_0}$$

C
$$\displaystyle\frac{1}{\sqrt{E_0}}$$

D
$$\displaystyle\sqrt{E_0}$$

Solution

Path traced by the electron is shown in the figure such that electron reaches the point P when its de Broglie wavelength becomes $$\dfrac{\lambda}{3}$$.
Acceleration of electron is zero in x direction.
Acceleration of electron in y direction $$a' = \dfrac{eE_o}{m}$$
Velocity of electron initially    $$v_o = \dfrac{h}{\lambda}$$
At P,   $$v_p = \dfrac{h}{(\lambda/3)} = 3v_o$$
So, velocity is y direction $$v' = \sqrt{(v')^2 - v_o^2} = \sqrt{9v_o^2-v_o^2} = 2\sqrt{2}v_o$$
Using   $$v' =a't$$
$$\therefore$$ $$2\sqrt{2}v_o = \dfrac{eE_o}{m}t$$
We get $$t = \dfrac{2\sqrt{2}v_o m}{eE_o}$$
$$\implies \ t\propto \dfrac{1}{E_o}$$
Question 3
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If the linear momentum of a particle is $$2.2 \times 10^4\, kg\, ms^{-1}$$, then what will be its de-Broglie wavelength ? (Take hr. $$6.6 \times 10^{-34}\, Js$$)

A
$$3 \times 10^{-29}\, m$$

B
$$3 \times 10^{-29}\, nm$$

C
$$6 \times 10^{-29}\, m$$

D
$$6 \times 10^{-29}\, nm$$

Solution

Given, the linear momentum of aparticular (p)
$$= 2.2 \times 10^4\, kg \, ms^{-1}$$
$$h= 6.6 \times 10^{-34} \, Js$$
The de-Broglie wavelength of particle
$$\lambda =\dfrac {h}{p}$$
$$\lambda =\dfrac {6.6 \times 10^{-34}}{2.2 \times 10^4}$$
or $$ \lambda = 3\times 10^{-38}\, m$$
or $$ \lambda = 3\times 10^{-29}\, nm$$
Question 4
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The wavelength $$\lambda$$ of a photon and the de-Broglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass m, speed of light c and planck constant.

A
$$\displaystyle\frac{\lambda mc}{h}$$

B
$$\displaystyle\frac{hmc}{\lambda}$$

C
$$\displaystyle\frac{2hmc}{\lambda}$$

D
$$\displaystyle\frac{2\lambda mc}{h}$$

Solution

The de-Broglie wavelength
$$h=\displaystyle\frac{h}{mv}\Rightarrow v=\frac{h}{m\lambda}$$ .........(i)
Energy of photon
$$E_p=\displaystyle\frac{hc}{\lambda}$$(since $$\lambda$$ is same) ...........(ii)
Energy of Photon
Kinetic energy of photon $$=\displaystyle\frac{E_p}{E_e}=\frac{hc/\lambda}{\displaystyle\frac{1}{2}1mu^2}=\frac{2hc}{\lambda mv^2}$$
Substituting value of v from Eq. (i), we get
$$\displaystyle\frac{E_p}{E_e}=\frac{2hc}{\lambda_m\left(\displaystyle\frac{h}{m\lambda}\right)^2}=\frac{2\lambda mc}{h}$$.
Question 5
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The energy of a $$K$$-electron in tungsten is $$-20keV$$ and of an $$L$$-electrons is $$-2keV$$. The wavelength of X-rays emitted when there is electron jump from $$L$$ to $$K$$ shell:

A
$$0.3443\mathring { A } $$

B
$$0.6887\mathring { A } $$

C
$$1.3982\mathring { A } $$

D
$$2.78\mathring { A } $$

Solution

A
$$E_{L} = - 2KeV =-2×10^{3}×1.6×10^{-19}J$$
$$E_{K} = - 20KeV =-20×10^{3}×1.6×10^{-19}J$$
The energy of the x-ray photon is given as
$$E = E_{L}-E_{K}= hv =\dfrac{hc}{\lambda }$$
Or, $$\lambda = \dfrac{hc}{E_{L}-E_{K}} =\dfrac{6.62×10^{-34}×3×10^{8}}{18×10^{3}1.6×10^{-19}}m = 0.6887\dot{A} $$
Question 6
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An electron and a photon have same wavelength of $$10^{-9}$$m. If E is the energy of the photon and p is the momentum of the electron, the magnitude of E/p in SI units is?

A
$$1.00\times 10^{-9}$$

B
$$1.50\times 10^8$$

C
$$3.00\times 10^8$$

D
$$1.20\times 10^7$$

Solution

Given:-
Wavelength of photon & electron are equal
$$ \Rightarrow { \lambda }_{ p }={ \lambda }_{ e }={ 10 }^{ -9 }m$$
$$ E$$=Energy of photon
$$ P$$= Momentum of electron
Solution:-
From Planck's equation
$$ E=\cfrac { hc }{ { \lambda }_{ p } } \Rightarrow { \lambda }_{ p }=\cfrac { hc }{ E } $$
From de-broglie's equation
$$ { \lambda }_{ e }=\cfrac { h }{ p }$$
Since $$ { \lambda }_{ p }={ \lambda }_{ e }\\ \Rightarrow \cfrac { hc }{ E } =\cfrac { h }{ p }$$
$$ \Rightarrow \cfrac { E }{ p } =C=3\times { 10 }^{ 8 }$$
Question 7
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A particle is projected horizontally with a velocity $$10\ m/s$$. What will be the ratio of de-Broglie wavelengths of the particle, when the velocity vector makes an angle $$30^{\circ}$$ and $$60^{\circ}$$ with the horizontal?

A
$$\sqrt {3} : 1$$

B
$$1 : \sqrt {3}$$

C
$$2 : \sqrt {3}$$

D
$$\sqrt {3} : 2$$

Solution

Given:-$${ V }_{ 1 }=10{ m }/{ s }$$

Solution:-

Velocity when velocity vector makes $${30}^{o}$$ with

horizontal,$${ V }_{ 2 }={ V }_{ 1 }\times \cos{ 30 }^{o}$$

$$ ={ V }_{ 1 }\times \cfrac { \sqrt { 3 } }{ 2 } $$

$${ V }_{ 2 }=\cfrac { 10\sqrt { 3 } }{ 2 } =5\sqrt { 3 } { m }/{ s }$$

Velocity when velocity vector makes $$ { 60 }^{o}$$ with

horizontal,$${ { V }_{ 3 } }={ V }_{ 1 }\cos{ 60 }^{ o }$$

$$ { V }_{ 3 }={ V }_{ 1 }\times \cfrac { 1 }{ 2 }$$

$${ V }_{ 3 }=10\times \cfrac { 1 }{ 2 } $$

$$ { V }_{ 3 }=5{ m }/{ s }$$

$$\therefore$$ Ratio of wavelength is:-

$$\sqrt { 3 } :1$$
Question 8
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A charged particle is accelerated from rest through a certain potential difference. The de Broglie wavelength is $${ \lambda }_{ 1 }$$ when it is accelerated through $${V}_{1}$$ and is $${ \lambda }_{ 2 }$$ when accelerated through $${V}_{2}$$. The ratio $${ \lambda }_{ 1 }/{ \lambda }_{ 2 }$$ is

A
$${ V }_{ 1 }^{ 3/2 }:{ V }_{ 2 }^{ 3/2 }$$

B
$${ V }_{ 2 }^{ 1/2 }:{ V }_{ 1 }^{ 1/2 }$$

C
$${ V }_{ 1 }^{ \cfrac { 1 }{ 2 } }:{ V }_{ 2 }^{ \cfrac { 1 }{ 2 } }$$

D
$${ V }_{ 1 }^{ 2 }:{ V }_{ 2 }$$

Solution

de Broglie wavelength is given by $$\lambda = \dfrac{h}{p}$$
where momentum $$ p = \sqrt{2mK} $$ and kinetic energy $$K =qV$$
$$\implies \ \lambda = \dfrac{h}{\sqrt{2mqV}}$$
We get $$\lambda \propto \dfrac{1}{\sqrt{V}}$$
$$\implies\ \lambda_1:\lambda_2 = V_2^{1/2}:V_1^{1/2}$$
Question 9
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An electron of mass m and a photon have same energy E.Find out the ratio of de-Brogile wavelength associated with them is (c- velocity of light)

A
$$c { \left[ 2mE \right] }^{ \dfrac { 1 }{ 2 } }$$

B
$$\dfrac { 1 }{ c } { \left[ \dfrac { 2m }{ E } \right] }^{ \dfrac { 1 }{ 2 } }$$

C
$$\dfrac { 1 }{ c } { \left[ \dfrac { E }{ 2m } \right] }^{ \dfrac { 1 }{ 2 } }$$

D
$${ \left[ \dfrac { E }{ 2m } \right] }^{ \dfrac { 1 }{ 2 } }$$

Solution
Question 10
1 / -0

Einstein's photoelectric equation is?

A
$$E_{max}=h\nu -\varphi_0$$

B
$$E_{max}=\displaystyle\frac{hc}{\lambda}+2\varphi_0$$

C
$$E_{max}=hv+\varphi_0$$

D
$$E_{max}=\displaystyle\frac{2hc}{\lambda}+\varphi_0$$

Solution

Photoelectrons from the metal surface are emitted when a light having frequency greater than the threshold frequency of the metal surface is incident on it.
Maximum kinetic energy of the emitted photo electrons is given by $$E_{max} = h\nu - \varphi_o$$
where $$\nu$$ is the frequency of incident light and $$\varphi_o$$ is the work function of metal surface.

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Dual Nature of Radiation and Matter Test - 40

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Dual Nature of Radiation and Matter Test - 40

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Question 1

$$4.24\times { 10 }^{ 6 }m/s\quad $$

$$5.25\times { 10 }^{ 6 }m/s$$

$$6.26\times { 10 }^{ 6 }m/s\quad $$

$$7.25\times { 10 }^{ 6 }m/s\quad $$

Solution

Question 2

$$E_0$$

$$\displaystyle\frac{1}{E_0}$$

$$\displaystyle\frac{1}{\sqrt{E_0}}$$

$$\displaystyle\sqrt{E_0}$$

Solution

Question 3

$$3 \times 10^{-29}\, m$$

$$3 \times 10^{-29}\, nm$$

$$6 \times 10^{-29}\, m$$

$$6 \times 10^{-29}\, nm$$

Solution

Question 4

$$\displaystyle\frac{\lambda mc}{h}$$

$$\displaystyle\frac{hmc}{\lambda}$$

$$\displaystyle\frac{2hmc}{\lambda}$$

$$\displaystyle\frac{2\lambda mc}{h}$$

Solution

Question 5

$$0.3443\mathring { A } $$

$$0.6887\mathring { A } $$

$$1.3982\mathring { A } $$

$$2.78\mathring { A } $$

Solution

Question 6

$$1.00\times 10^{-9}$$

$$1.50\times 10^8$$

$$3.00\times 10^8$$

$$1.20\times 10^7$$

Solution

Question 7

$$\sqrt {3} : 1$$

$$1 : \sqrt {3}$$

$$2 : \sqrt {3}$$

$$\sqrt {3} : 2$$

Solution

Question 8

$${ V }_{ 1 }^{ 3/2 }:{ V }_{ 2 }^{ 3/2 }$$

$${ V }_{ 2 }^{ 1/2 }:{ V }_{ 1 }^{ 1/2 }$$

$${ V }_{ 1 }^{ \cfrac { 1 }{ 2 } }:{ V }_{ 2 }^{ \cfrac { 1 }{ 2 } }$$

$${ V }_{ 1 }^{ 2 }:{ V }_{ 2 }$$

Solution

Question 9

$$c { \left[ 2mE \right] }^{ \dfrac { 1 }{ 2 } }$$

$$\dfrac { 1 }{ c } { \left[ \dfrac { 2m }{ E } \right] }^{ \dfrac { 1 }{ 2 } }$$

$$\dfrac { 1 }{ c } { \left[ \dfrac { E }{ 2m } \right] }^{ \dfrac { 1 }{ 2 } }$$

$${ \left[ \dfrac { E }{ 2m } \right] }^{ \dfrac { 1 }{ 2 } }$$

Solution

Question 10

$$E_{max}=h\nu -\varphi_0$$

$$E_{max}=\displaystyle\frac{hc}{\lambda}+2\varphi_0$$

$$E_{max}=hv+\varphi_0$$

$$E_{max}=\displaystyle\frac{2hc}{\lambda}+\varphi_0$$

Solution

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