Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

A particle $$A$$ of mass $$m$$ and initial velocity $$v$$ collides with a particle $$B$$ of mass $$\cfrac{m}{2}$$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $${ \lambda }_{ A }$$ and $${ \lambda }_{ B }$$ after the collision is :

A
$$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 1 }{ 2 } $$

B
$$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 1 }{ 3 } $$

C
$$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =2$$

D
$$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 2 }{ 3 } $$

Solution

$$mv=m{ v }_{ A }+\cfrac { m }{ 2 } { v }_{ B }$$ (conservation of linear momentum)
$$\quad \therefore v={ v }_{ A }+\cfrac { { v }_{ B } }{ 2 } ={ v }_{ B }-{ v }_{ A }\quad $$ (elastic collision)
$$\therefore \cfrac { { v }_{ B } }{ { v }_{ A } } =4\quad \quad \therefore \cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { { { m }_{ B }v }_{ B } }{ { { m }_{ A }v }_{ A } } =2$$
Question 2
1 / -0

The de Broglie wavelength of an electron in a metal at $$27^{\circ}C$$ is $$(Given \, m_e \, = \, 9.1 \, \times \, 10^{-31} \, kg,\, k_B \, = \, 1.38 \, \times \, 10^{-23} \, J \, K^{-1})$$

A
$$6.2 \, \times \, 10^{-9} \, m$$

B
$$6.2 \, \times \, 10^{-10} \, m$$

C
$$6.2 \, \times \, 10^{-8} \, m$$

D
$$6.2 \, \times \, 10^{-7} \, m$$

Solution

Here, T = 27 + 273 = 300 K
For an electron in a metal, momentum $$p = \sqrt{3mk_{B}T}$$
de Broglie wavelength of an electron is
$$\lambda \, = \dfrac{h}{p} \, = \dfrac{h}{\sqrt{3mk_{B}T}} $$
$$= \dfrac{h}{\sqrt{3 \times \left ( 9.1 \times 10^{-31} \right ) \times \left ( 1.38 \times 10^{-23} \right )\times 300}} $$

$$= \, 6.2 \times 10^{-9} m $$
Question 3
1 / -0

Two large vertical and parallel metal plates having a separation of $$1\ cm$$ are connected to a $$DC$$ voltage source of potential difference $$X$$. A proton is released at rest midway between the two plates. It is found to move at $${45}^{o}$$ to the vertical JUST after release. Then $$X$$ is nearly

A
$$1\times { 10 }^{ -5 }V$$

B
$$1\times { 10 }^{ -7 }V$$

C
$$1\times { 10 }^{ -9 }V$$

D
$$1\times { 10 }^{ -10 }V$$

Solution

For proton to move at $$45^{o}$$
$$qE= mg$$
so that the net force is along $$45^{o}$$
Now $$qE= mg$$ and $$E= \dfrac{V}{d}$$
$$\dfrac{1.6}{10^{19}} \times \dfrac{V \times 100}{1}= \dfrac{1.6}{10^{27}} \times 10$$
$$V= 1 \times 10^{-9} V$$
Question 4
1 / -0

An atom of nuclean change $$q$$ is placed on $$a$$ line. There particles $$A (mass = m, change = q), B(mass = 2, change = q)$$ and $$C (mass = 3m, change = 3q)$$ are shot towards the atom, a distance $$r$$ away from the line and parallel to it. The particle that would be deflected the most would be?

A
$$A$$

B
$$B$$

C
$$C$$

D
All have the same deflection
Question 5
1 / -0

A radiation of energy $$E$$ falls normally on a perfectly reflecting surface. The momentun transferred to the surface is

A
$$E/c$$

B
$$2E/c$$

C
$$Ec$$

D
$$E/{c}^{2}$$

Solution

Let initial momentum is $${{p}_{i}}$$and final reflected momentum is $${{p}_{f}}$$such that

Energy, $$E=\dfrac{hc}{\lambda }.....(1)$$

$$ {{p}_{i}}=\dfrac{h}{\lambda }=\dfrac{E}{c}\,\,(from\,\,1) $$

$$ {{p}_{f}}=\dfrac{-h}{\lambda }=\dfrac{-E}{c} $$

So, net change in momentum is

$$ \Delta p={{p}_{f}}-{{p}_{i}} $$

$$ \Delta p=\dfrac{-E}{c}-\dfrac{E}{c} $$

$$ \Delta p=-\dfrac{2E}{c} $$

The momentum transferred to the surface is$$\dfrac{2E}{c}$$
Question 6
1 / -0

The uncertainty in position of a particle is same as it's de Broglie wavelength, uncertainty in its momentum is _______

A
$$\cfrac { h }{ \lambda } $$

B
$$\cfrac { \lambda }{ h } $$

C
$$\cfrac { 2h }{ 3\lambda } $$

D
$$\cfrac { 3\lambda }{ 2h } $$

Solution

$$\Delta x=\lambda$$ $$\left( \cfrac { a }{ q } \right) $$
According to the Uncertainty principle,
$$\Delta x\cdot \Delta p\ge \cfrac { h }{ 4\pi } $$
$$\therefore \Delta p\ge \cfrac { 1 }{ 4\pi } \left( \cfrac { h }{ \lambda } \right) \\ \therefore \Delta p\propto \cfrac { h }{ \lambda } $$
Question 7
1 / -0

Electrons are accelerated through a potential difference $$V$$ and protons are accelerated through a potential difference $$4V$$. The de-Broglie wavelengths are $${ \lambda }_{ e }$$ and $${ \lambda }_{ p }$$ for electrons are protons respectively. The ratio of $$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } $$ is given by (given $${m}_{e}$$ is mass of electron and $${m}_{p}$$ is mass of proton)

A
$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =2\sqrt { \cfrac { { m }_{ p } }{ { m }_{ e } } } $$

B
$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } } $$

C
$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\cfrac { 1 }{ 2 } \sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } } $$

D
$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =2\sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } } $$

Solution

$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\cfrac { \sqrt { 2{ m }_{ p }{ k }_{ p } } }{ \sqrt { 2{ m }_{ e }{ k }_{ e } } } =\cfrac { \sqrt { 2\times { m }_{ p }\times e\times 4v } }{ \sqrt { 2\times { m }_{ e }\times e\times v } } \Rightarrow \cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =2\sqrt { \cfrac { { m }_{ p } }{ { m }_{ e } } } $$
Question 8
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In a photoelectric experiment anode potential is plotted against plate current. Then:

A
$$A$$ and $$B$$ will have different intensities while $$B$$ and $$C$$ will have different frequencies.

B
$$B$$ and $$C$$ will have different intensities while $$A$$ and $$C$$ will have different frequencies.

C
$$A$$ and $$B$$ will have different intensities while $$A$$ and $$D$$ will have different frequencies.

D
$$A$$ and $$B$$ will have different intensities while $$B$$ and $$C$$ will have same frequencies.

Solution
Question 9
1 / -0

According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal $${V}_{s}$$ the frequency, of the incident radiation gives straight line whose slope

A
depends on the nature of the metal used

B
depends on the intensity of the radiation

C
depends both on the intensity of the radiation and the metal used

D
is the same for all metals and independent of the intensity of the radiation

Solution

Einstein's photoelectric equation is $${ KE }_{ max }=hc-\phi \quad \longrightarrow \left( i \right) $$
The equation of the line is $$y=mx+C\quad \longrightarrow \left( ii \right) $$
$$m=h$$, $$C=-\phi $$
Question 10
1 / -0

The audio signal:

A
can be sent directly over the air for large distance

B
cannot be sent directly over the air for large distance

C
posses very high frequency

D
none of above

Solution

An Audio signal may vary from $$20Hz$$ to $$20KHz$$,
But in general, it is around $$200Hz$$ only,.
This Frequency when transmitted directly to air for long-distance is attenuated
and a signal is lost in the air, That's why such frequency is first modulated to higher frequencies of order(MHz) and then transmitted to air.
Hence audio signals can't be transmitted directly to air for long distances.

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Dual Nature of Radiation and Matter Test - 43

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Dual Nature of Radiation and Matter Test - 43

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Question 1

$$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 1 }{ 2 } $$

$$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 1 }{ 3 } $$

$$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =2$$

$$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 2 }{ 3 } $$

Solution

Question 2

$$6.2 \, \times \, 10^{-9} \, m$$

$$6.2 \, \times \, 10^{-10} \, m$$

$$6.2 \, \times \, 10^{-8} \, m$$

$$6.2 \, \times \, 10^{-7} \, m$$

Solution

Question 3

$$1\times { 10 }^{ -5 }V$$

$$1\times { 10 }^{ -7 }V$$

$$1\times { 10 }^{ -9 }V$$

$$1\times { 10 }^{ -10 }V$$

Solution

Question 4

$$A$$

$$B$$

$$C$$

All have the same deflection

Question 5

$$E/c$$

$$2E/c$$

$$Ec$$

$$E/{c}^{2}$$

Solution

Question 6

$$\cfrac { h }{ \lambda } $$

$$\cfrac { \lambda }{ h } $$

$$\cfrac { 2h }{ 3\lambda } $$

$$\cfrac { 3\lambda }{ 2h } $$

Solution

Question 7

$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =2\sqrt { \cfrac { { m }_{ p } }{ { m }_{ e } } } $$

$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } } $$

$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\cfrac { 1 }{ 2 } \sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } } $$

$$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =2\sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } } $$

Solution

Question 8

$$A$$ and $$B$$ will have different intensities while $$B$$ and $$C$$ will have different frequencies.

$$B$$ and $$C$$ will have different intensities while $$A$$ and $$C$$ will have different frequencies.

$$A$$ and $$B$$ will have different intensities while $$A$$ and $$D$$ will have different frequencies.

$$A$$ and $$B$$ will have different intensities while $$B$$ and $$C$$ will have same frequencies.

Solution

Question 9

depends on the nature of the metal used

depends on the intensity of the radiation

depends both on the intensity of the radiation and the metal used

is the same for all metals and independent of the intensity of the radiation

Solution

Question 10

can be sent directly over the air for large distance

cannot be sent directly over the air for large distance

posses very high frequency

none of above

Solution

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