Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 53

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Question 1
1 / -0

The de-Broglie wavelength $$(\lambda _B)$$ associated with the electron orbiting in the second excited state of hydrogen atom is relared to that in the ground state $$(\lambda _G)$$ by:

A
$$(\lambda _B=3 {\lambda _G})$$

B
$$(\lambda _B=2 {\lambda _G})$$

C
$$(\lambda _B=3 {\lambda _{G/3}})$$

D
$$(\lambda _B=3 {\lambda _{G/2}})$$
Question 2
1 / -0

If the deBroglie wavelenght of an electron is equal to $$10^{3}$$ times the wavelength of a photon of frequency $$6 \times 10^{14} Hz$$, then the speed of electron is equal to : (Speed of light = $$3 \times 10^8 m/s$$ Planck's constant = $$6.63 \times 10^{34} J$$ . Mass of electron = $$9.1 10^{31} kg$$)

A
$$1.45\times 10^6m/s$$

B
$$1.75\times 10^6m/s$$

C
$$1.8\times 10^6m/s$$

D
$$1.1\times 10^6 m/s$$

Solution

$$\dfrac{h}{mv}=10^{-3}\left(\dfrac{3\times 10^8}{6\times 10^{14}}\right)$$
$$v=\dfrac{6.63\times 10^{-34}\times 6\times 10^{14}}{9.1\times 10^{-31}\times 3\times 10^5}$$
Question 3
1 / -0

If there is an increase in linear dimensions of the objects, the associated de-broglie wavelength.

A
Increases

B
Decreases

C
Remains unchanged

D
Depends on the density

Solution

As we know,
By de Broglie wavelength equation,
Wavelength $$=$$ plank's constant $$\div $$ linear moment
i.e. wavelength $$\left( h \right) \alpha \dfrac { 1 }{ { \left( P \right) }_{ momentum } } $$
If mass is increases momentum increases than de Broglie wavelength decreases.
Question 4
1 / -0

A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. if the speed of the electrons is increased , which of the following statements is correct?

A
The angular width of the central maximum will decrease.

B
The angular width of the central maximum will be unaffacted.

C
Diffraction pattern is not observed on the screen in the case of electrons.

D
The angular width of the central maximum of the diffraction pattern will increase.
Question 5
1 / -0

If velocity of a particle is 3 times of that of electron and ratio of de brogile wavelength of particle to that of electron is $$1.814\times { 10 }^{ -4 }.$$ The particle will be:-

A
Neutron

B
Deutron

C
Alpha

D
Tritium

Solution

We know that,
$$\dfrac { { m }_{ p } }{ { m }_{ e } } =\dfrac { 1.7\times { 10 }^{ -27 } }{ 9.1\times { 10 }^{ -31 } } $$
$$\dfrac { { m }_{ p } }{ { m }_{ e } } $$ for alpha particle is -
Mass of $$1$$ mole of $$\alpha$$ particle $$=4g/mol$$
Mass of $$1\alpha $$ particle $$=\dfrac { 4 }{ 6.022\times { 10 }^{ -23 } } =6.64\times { 10 }^{ -27 }kg$$
$${ m }_{ p }/{ m }_{ e }$$ for alpha particles $$=\dfrac { 6.64\times { 10 }^{ -24 } }{ 9.1\times { 10 }^{ -21 } } =7300$$
$${ m }_{ p }/{ m }_{ e }$$ for deuterium particle $$=3650$$
$${ m }_{ p }/{ m }_{ e }$$ for Tritium particle $$=\dfrac { 9.98\times { 10 }^{ -27 } }{ 9.1\times { 10 }^{ -31 } } =57.70$$
$$\therefore$$ $${ m }_{ p }/{ m }_{ e }$$ ratio for neutron's is closest.
Question 6
1 / -0

A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values $${ \lambda }_{ 1 }$$ and $${ \lambda }_{ 2 }$$ with $${ \lambda }_{ 1 }>{ \lambda }_{ 2 }.$$ Which of the following statement is true?

A
The particle could be moving in an circular orbit with origin as centre

B
The particle could be moving in an parabolic orbit with origin as its focus

C
When the de Broglie wave length is $${ \lambda }_{ 1 },$$ the particle is nearer the origin than when its value is $${ \lambda }_{ 2 }$$.

D
When the de Broglie wavelength is $${ \lambda }_{ 2 },$$ the particle is nearer the origin than when its value is $${ \lambda }_{ 1 }$$.

Solution

We know that,
$${ \lambda }_{ 1 }=\dfrac { h }{ { mv }_{ 1 } } $$ and $${ \lambda }_{ 2 }=\dfrac { h }{ { mv }_{ 2 } } $$
$$\therefore$$ $$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 1 } } =\dfrac { { v }_{ 2 } }{ { v }_{ 1 } } $$ and thus $${ \lambda }_{ 1 }>{ \lambda }_{ 2 }$$
By lower of conversation of angular momentum the particle moves faster when it is closer to focus.
Question 7
1 / -0

During an experiment an $$\alpha $$-particle and a proton are accelerated by same positive difference , their de broglie wavelength ratio will (Take mass of proton= mass of neutron).

A
$$1:2$$

B
$$1:4$$

C
$$1:2\sqrt { 2 } $$

D
$$1:\sqrt { 2 } $$

Solution
Question 8
1 / -0

You are given an electron with a deBroglie wavelength of $$\lambda =76.3$$ nm. What is the Kelvin temperature of this electron ?

A
1.50

B
2.00

C
2.50

D
3.00

Solution

$$\begin{array}{l} \lambda =\dfrac { h }{ p } \, \, and\, \, \, \dfrac { { { p^{ 2 } } } }{ { 2m } } =\varepsilon =\dfrac { 3 }{ 2 } kT \\ \therefore p={ \left( { 3mkT } \right) ^{ \dfrac { 1 }{ 2 } } } \\ \lambda =\dfrac { h }{ p } =\dfrac { h }{ { { { \left( { 3kmT } \right) }^{ \dfrac { 1 }{ 2 } } } } } \\ \therefore T=\dfrac { { { h^{ 2 } } } }{ { 3km{ \lambda ^{ 2 } } } } \\ T=\dfrac { { { h^{ 2 } } } }{ { 3k{ \lambda ^{ 2 } } } } =\dfrac { { { { \left( { 6.626\times { { 10 }^{ -34 } } } \right) }^{ 2 } } } }{ { \left( 3 \right) \left( { 1.38\times { { 10 }^{ -23 } } } \right) \left( { 9.108\times { { 10 }^{ -31 } } } \right) { { \left( { 76.3\times { { 10 }^{ -9 } } } \right) }^{ 2 } } } } =2.00K \\ T=2.00K \end{array}$$

Option $$B$$ is correct .
Question 9
1 / -0

In a photoelectron experiment, the stopping potential from the photoelectrons is $$2V$$ for the incident of wavelength $$400 nm$$. If the incident light is changed to $$300 nm$$, the cut-off potential is:

A
$$2V$$

B
Greater than $$8/3V$$

C
$$8/3V$$

D
Zero
Question 10
1 / -0

The K.E. of electron and photon is same then the relation between their De- Broglie wavelength:

A
$${\lambda _p} < {\lambda _e}$$

B
$${\lambda _p} = {\lambda _e}$$

C
$${\lambda _p} > {\lambda _e}$$

D
$${\lambda _p} = 2{\lambda _e}$$

Solution

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Dual Nature of Radiation and Matter Test - 53

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Dual Nature of Radiation and Matter Test - 53

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Question 1

$$(\lambda _B=3 {\lambda _G})$$

$$(\lambda _B=2 {\lambda _G})$$

$$(\lambda _B=3 {\lambda _{G/3}})$$

$$(\lambda _B=3 {\lambda _{G/2}})$$

Question 2

$$1.45\times 10^6m/s$$

$$1.75\times 10^6m/s$$

$$1.8\times 10^6m/s$$

$$1.1\times 10^6 m/s$$

Solution

Question 3

Increases

Decreases

Remains unchanged

Depends on the density

Solution

Question 4

The angular width of the central maximum will decrease.

The angular width of the central maximum will be unaffacted.

Diffraction pattern is not observed on the screen in the case of electrons.

The angular width of the central maximum of the diffraction pattern will increase.

Question 5

Neutron

Deutron

Alpha

Tritium

Solution

Question 6

The particle could be moving in an circular orbit with origin as centre

The particle could be moving in an parabolic orbit with origin as its focus

When the de Broglie wave length is $${ \lambda }_{ 1 },$$ the particle is nearer the origin than when its value is $${ \lambda }_{ 2 }$$.

When the de Broglie wavelength is $${ \lambda }_{ 2 },$$ the particle is nearer the origin than when its value is $${ \lambda }_{ 1 }$$.

Solution

Question 7

$$1:2$$

$$1:4$$

$$1:2\sqrt { 2 } $$

$$1:\sqrt { 2 } $$

Solution

Question 8

1.50

2.00

2.50

3.00

Solution

Question 9

$$2V$$

Greater than $$8/3V$$

$$8/3V$$

Zero

Question 10

$${\lambda _p} < {\lambda _e}$$

$${\lambda _p} = {\lambda _e}$$

$${\lambda _p} > {\lambda _e}$$

$${\lambda _p} = 2{\lambda _e}$$

Solution

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