Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Although we can know with great precision the pattern formed by huge numbers of photons passing through the lines on a diffraction grating, we cannot know at all where one single photon will go when it passes through the diffraction grating.
This statement describes the truth revealed by what theory?

A
Universal gravitation

B
Chaos theory

C
Relativity

D
Quantum theory

E
Group theory

Solution

According to quantum theory of light, light has dual nature, it acts as a wave in case of diffraction of waves.

But light also acts as a particle. It is made up of quanta called photons. Each photon when falls on a material, it ejects one electron with energy $$ E = h\nu $$
So, because of wave nature of light we can see the diffraction pattern formed by light but because of particle nature of light, where one single photon goes as it passes through diffraction grating cannot be determined.
Question 2
1 / -0

A photo cell is a device which

A
Absorbs light and produces a stream of electrons

B
Absorb a stream of electron and produces light

C
Converts protons into photons

D
Converts photons into protons

Solution

Photocell is a device which converts light energy into electrical energy. In photocell concave metallic cathode absorbs photons from incident light waves and emits electron from the metal surface towards anode.
In other words, it absorbs light and produces a stream of electrons.
Hence, option A is correct.
Question 3
1 / -0

Which of the following statement is correct?

A
Photo-current increases with intensity of light

B
Photo-current is proportional to the applied voltage

C
Current in photocell increases with increasing frequency

D
Stopping potential increases with increase of incident light

Solution

More is the intensity of incident radiation, more is the umber of photons and thus more is the number of photo electrons emitted. This results in the increase in the photo-current.
Question 4
1 / -0

The de-Broglie wavelength associated with a particle with rest mass $$m_0$$ and moving with speed of light in vacuum is ___________.

A
$$\dfrac{h}{m_0c}$$

B
$$0$$

C
$$\infty$$

D
$$\dfrac{m_0c}{h}$$

Solution

De-Broglie wavelength $$\lambda=\dfrac{h}{p}$$
where $$h=$$Plank's constant
$$p=$$ momentum associated with a particle
Given:
$$mass=m_o$$
$$speed=c$$
Therefore $$p=m_oc$$
Hence $$\lambda=\dfrac{h}{p}=\dfrac{h}{m_oc}$$
Therefore the correct option is (A).
Question 5
1 / -0

If the wavelength is brought down from $$6000 \mathring{A}$$ to $$4000 \mathring{A}$$ in a photoelectric experiment, then what will happen ?

A
The work function of the metal will increase

B
The threshold frequency will decrease

C
No change will take place

D
Cut off voltage will increase

Solution

We know,
Stopping potential (or cut off voltage) is given by $$V_o=\dfrac{hc}{e}(\dfrac{1}{\lambda})-\dfrac{\phi}{e}$$
where $$\lambda$$ is the wavelength of the incident photon.
Since work function $$\phi$$ is the property of the metal surface, thus it does not depend on the wavelength of incident photon. Hence $$\phi$$ remains the same.
As per the equation,
when the wavelength of the incident photon decreases, the stopping potential will increase according to the equation.
Question 6
1 / -0

The de-Broglie wavelength associated with thermal neutron ( t=51C say) is of the order of the

A
distance between the atoms in the crystal

B
size of the nucleus

C
Bohr's radius

D
size of grain

Solution

$$ T = 51^\circ C $$
So, temp in Kelvin = $$ 51 + 273 = 324 K $$
de-broglie wavelength $$ \lambda = \dfrac{h}{p} $$
$$ p = \sqrt{2mE} $$

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

$$ m $$ of neutron $$ = 1.67 \times 10^{-27} kg$$

Also, $$ E = \dfrac{3}{2} kT$$
where $$ k = \text{Boltzman constant} = 1.38 \times 10^{-23} J \, mol^{-1} \, K^{-1} $$

So, $$ \lambda = \dfrac{h}{\sqrt{3mkT}}$$

Substituting the values,

$$ \lambda = 1.39 \times 10^{10} m$$
which is the order of distance between atoms in the crystal.
Question 7
1 / -0

A proton is accelerated to $$225V$$. Its de-Broglie wavelength is:

A
$$0.0019nm$$

B
$$0.02nm$$

C
$$0.003nm$$

D
$$0.4nm$$

Solution

Energy of proton = $$ \cfrac{p^2}{2m} = qV$$
$$ \cfrac{p^2}{2m} = 225eV$$
$$ p = \sqrt{2m \times 225 \times 1.6 \times 10^{-19}}$$
$$ \therefore \lambda = \cfrac{h}{p}$$
$$ = \cfrac{6.626 \times 10^{-34}}{\sqrt{2m \times 225 \times 1.6 \times 10^{-19}}}$$
$$ = 0.19 \times 10^{-11} m$$
$$ = 1.9 \times 10^{-12} m $$
$$ = 1.9 pm$$ or $$ 0.001 nm$$
Question 8
1 / -0

De-Broglie wavelength of an atom at absolute temperature $$T\ K$$ will be

A
$$\dfrac {h}{\sqrt {3mKT}}$$

B
$$\dfrac {h}{mKT}$$

C
$$\dfrac {\sqrt {2mKT}}{h}$$

D
$$\sqrt {2mKT}$$

Solution

Key Concept The average kinetic energy of atom at temperature $$T\ K$$ is
$$E = \dfrac {3}{2}KT$$
Now, kinetic energy of an atom $$E = \dfrac {P^{2}}{2m}$$
$$\therefore P = \sqrt {2ME}$$
$$\therefore$$ de-Broglie wavelength
$$\lambda = \dfrac {h}{P} = \dfrac {h}{\sqrt {2mE}}$$
$$\Rightarrow \lambda = \dfrac {h}{\sqrt {2m\times \dfrac {3}{2}KT}} = \dfrac {h}{\sqrt {3mKT}}$$.
Question 9
1 / -0

If the mass of a microscopic particle a well as its speed are halved, the de-broglie wavelength associated with the particle will

A
increased by a factor more than $$2$$

B
increases by a factor of $$2$$

C
decreases by a factor of $$2$$

D
decrease by a factor more than $$2$$

Solution

$$ \lambda $$ is directly proportional to $$ \dfrac{1}{mv}$$

$$\dfrac{ \lambda_1}{ \lambda_2} = \dfrac{m_2 v_2}{ m_1 v_1}$$

$$ \dfrac{ \lambda_1}{ \lambda_2} = \dfrac{1}{2} \times \dfrac{1}{2}$$

$$ \lambda_2 = 4 \lambda_1$$
Question 10
1 / -0

If a proton and electron have the same de-Broglie wavelength, then

A
Kinetic energy of electron < kinetic energy of proton

B
Kinetic energy of electron = kinetic energy of proton

C
Momentum of electron > momentum of proton

D
Momentum of electron = momentum of proton

E
Momentum of electron < momentum of proton

Solution

de-Broglie wavelength is given by $$\lambda=\dfrac{h}{p}$$,
Where $$h$$ is $$Planck's $$ constant so $$same $$ for both electron and proton.
$$p$$ is $$momentum $$ of the particle, as both have same wavelength so $$momentum$$ should be $$same$$ for both of them.

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Dual Nature of Radiation and Matter Test - 21

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Dual Nature of Radiation and Matter Test - 21

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Question 1

Universal gravitation

Chaos theory

Relativity

Quantum theory

Group theory

Solution

Question 2

Absorbs light and produces a stream of electrons

Absorb a stream of electron and produces light

Converts protons into photons

Converts photons into protons

Solution

Question 3

Photo-current increases with intensity of light

Photo-current is proportional to the applied voltage

Current in photocell increases with increasing frequency

Stopping potential increases with increase of incident light

Solution

Question 4

$$\dfrac{h}{m_0c}$$

$$0$$

$$\infty$$

$$\dfrac{m_0c}{h}$$

Solution

Question 5

The work function of the metal will increase

The threshold frequency will decrease

No change will take place

Cut off voltage will increase

Solution

Question 6

distance between the atoms in the crystal

size of the nucleus

Bohr's radius

size of grain

Solution

Question 7

$$0.0019nm$$

$$0.02nm$$

$$0.003nm$$

$$0.4nm$$

Solution

Question 8

$$\dfrac {h}{\sqrt {3mKT}}$$

$$\dfrac {h}{mKT}$$

$$\dfrac {\sqrt {2mKT}}{h}$$

$$\sqrt {2mKT}$$

Solution

Question 9

increased by a factor more than $$2$$

increases by a factor of $$2$$

decreases by a factor of $$2$$

decrease by a factor more than $$2$$

Solution

Question 10

Kinetic energy of electron < kinetic energy of proton

Kinetic energy of electron = kinetic energy of proton

Momentum of electron > momentum of proton

Momentum of electron = momentum of proton

Momentum of electron < momentum of proton

Solution

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