Dual Nature of Radiation and Matter Test

Result

Dual Nature of Radiation and Matter Test - 16

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A proton when accelerated through a potential difference of V volt has a wavelength $$\lambda $$ associated with it. An $$\alpha $$ - particle in order to have the same wavelength $$\lambda $$ must be accelerated through a p.d. of

A
V/8 volt

B
V/4 volt

C
V volt

D
2V volt

Solution

For proton, $$\lambda_p =\dfrac{0.286\times 10^{-10}}{\sqrt{\nu }}\ m$$
where V is p.d
For $$\alpha -particle,\ \lambda_\alpha =\dfrac{0.101\times 10^{-10}}{\sqrt{\nu }}m$$

so, keeping same $$\alpha $$ for proton and $$\alpha -partside$$

$$1= \dfrac{\lambda _p}{\lambda _{\alpha }} = \dfrac{0.286 \times \sqrt{V_{\alpha }}}{0.101 \times \sqrt{V_{P}}}$$

$$0.101 \times \sqrt{V_{P}} = 0.286 \times \sqrt{V_{\alpha }}$$
$$(0.101)^{2} \times V_{P} = (0.286)^{2}V _{\alpha} $$

$$\dfrac{(0.101)^{2}\nu }{(0.286)^{2}} = V _{\alpha }$$

$$(\because V_{P}=V)$$

$$\Rightarrow V_{\alpha } = \dfrac{V}{8} volt$$

So, the answer is option (A).
Question 2
1 / -0

Moving with the same velocity, one of the following has the longest de Broglie wavelength

A
$$\beta -$$particle

B
$$\alpha -$$particle

C
proton

D
neutron

Solution

De-broglie wavelength
$$\lambda =\dfrac{h}{mv}$$
$$\lambda \propto \dfrac{1}{m}$$
$$\beta$$ particle has least mass.
So $$\beta$$ particle has the longest wavelength.
Question 3
1 / -0

The wavelength associated with an electron having kinetic energy is given by the expression:

A
$$h/\sqrt{2mE}$$

B
$$2h/mE$$

C
$$2 mhE$$

D
$$\dfrac{2\sqrt{2mE}}{h}$$

Solution

$$\dfrac{1}{2}mv^{2}=E\ (given)$$

$$mv=\sqrt{2Em}$$

$$\therefore \ \lambda =\dfrac{h}{mv}$$

$$=\dfrac{h}{\sqrt{2mE}}$$
So, the answer is option (A).
Question 4
1 / -0

De Broglie wavelength ‘$$\lambda$$ ’ is proportional to

A
$$\dfrac{1}{\sqrt{E}}$$ for photons and $$\dfrac{1}{E}$$ for particles

B
$$\dfrac{1}{E}$$ for photons and $$\dfrac{1}{\sqrt{E}}$$ for particles

C
$$\dfrac{1}{E}$$ for both photons and particles in motion

D
$$\dfrac{1}{\sqrt{E}}$$ for both photons and particles

Solution

For photon
$$E=\dfrac{hc}{\lambda }$$

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

$$\therefore \ \lambda \ \alpha \dfrac{1}{E}$$

and for partides
$$\dfrac{1}{2}mv^{2} = E$$

$$mv = \sqrt{2Em}$$

$$\therefore \lambda = \dfrac{h}{mv}$$

$$=\dfrac{h}{\sqrt{2Em}}$$

$$\lambda \alpha \dfrac{1}{\sqrt{E}}$$
So, the answer is option (B).
Question 5
1 / -0

A microscopic particle of mass $$10^{-12}$$ kg is moving with a velocity of $$10^{2}$$ m/s. Then the de-Broglie wavelength associated with the particle is

A
$$6.6\times 10^{-24}m$$

B
$$6.6\times 10^{34}m$$

C
$$3.3\times 10^{24}m$$

D
$$0$$

Solution

$$\lambda =\dfrac{h}{m\nu }$$

$$=\dfrac{6.6\times 10^{-34}}{10^{-12}\times 10^{2}}$$

$$=6.6\times 10^{-24}m$$
So, the answer is option (A).
Question 6
1 / -0

An electron accelerated under a p.d. of V volt has a certain wavelength $$\lambda$$ . Mass of the proton is 2000 times the mass of an electron. If the proton has to have the same wavelength $$\lambda$$ , then it will have to be accelerated under p.d. of (volts):

A
$$100$$

B
$$2000$$

C
$$V/2000$$

D
$$\sqrt{2000}$$

Solution

$$\dfrac{1}{2}mv^{2}=eV$$
$$mv=\ \sqrt{2meV}$$
$$Now \ \lambda _{e}=\dfrac{h}{mv}$$
$$=\dfrac{h}{\sqrt{2meV}}$$
$$\lambda_p =\ \dfrac{h}{\sqrt{2\times 2000m\times e\times V^{1}}}$$
$$Given,\ \lambda _{e} = \lambda _p$$
$$V^{1} = \dfrac{V}{2000}$$
Question 7
1 / -0

The value of de Broglie wavelength of an electron moving with a speed of $$6.6 \times 10^{5} ms^{-1}$$ is approximately

A
11 $$A^{0}$$

B
111$$A^{0}$$

C
211 $$A^{0}$$

D
311 $$A^{0}$$

Solution

$$\lambda =\dfrac{h}{m\nu }$$

$$=\dfrac{6.6\times 10^{-34}}{9.1\times 10^{-31}\times 6.6\times 10^{5}}$$

$$=11\ A^{\circ}$$
So, the answer is option (A).
Question 8
1 / -0

The de Broglie wavelength associated with a particle at rest is

A
0

B
$$\infty $$

C
finite

D
data insufficient

Solution

The relationship between de Broglie wavelength $$(\lambda)$$ and the momentum $$(p)$$ of a photon is given by:
$$\lambda=\dfrac{h}{p}$$. So, the de Broglie wavelength associated with a particle at rest $$(p=0)$$ is infinity.
So, the answer is option (B).
Question 9
1 / -0

Matter waves are:

A
Electromagnetic waves

B
Mechanical waves

C
Either mechanical or electromagnetic waves

D
Neither mechanical nor electromagnetic waves

Solution

Correct Answer: option (D)
Matter waves are waves associated with every moving particle and therefore are neither mechanical nor electromagnetic waves.
These waves have
$$\lambda = h\times p$$
Where $$\lambda$$ is the wavelength, $$h$$ is plank's constant, and $$p$$ is the particle’s moment.
Therefore, matter waves are neither mechanical nor electromagnetic waves.
Question 10
1 / -0

Einstein was awarded a Noble Prize for :

A
Photo electric effect

B
Compton effect

C
Theory of relativity

D
None of the above

Solution

The noble prize in Physics $$1921$$ was awarded to Albert Einstein for his services to theoretical Physics and especially for his discovery of the two of the photoelectric effect

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Dual Nature of Radiation and Matter Test - 16

Result

Dual Nature of Radiation and Matter Test - 16

Score

-

Rank

-

SHARING IS CARING

Question 1

V/8 volt

V/4 volt

V volt

2V volt

Solution

Question 2

$$\beta -$$particle

$$\alpha -$$particle

proton

neutron

Solution

Question 3

$$h/\sqrt{2mE}$$

$$2h/mE$$

$$2 mhE$$

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

Solution

Question 4

$$\dfrac{1}{\sqrt{E}}$$ for photons and $$\dfrac{1}{E}$$ for particles

$$\dfrac{1}{E}$$ for photons and $$\dfrac{1}{\sqrt{E}}$$ for particles

$$\dfrac{1}{E}$$ for both photons and particles in motion

$$\dfrac{1}{\sqrt{E}}$$ for both photons and particles

Solution

Question 5

$$6.6\times 10^{-24}m$$

$$6.6\times 10^{34}m$$

$$3.3\times 10^{24}m$$

$$0$$

Solution

Question 6

$$100$$

$$2000$$

$$V/2000$$

$$\sqrt{2000}$$

Solution

Question 7

11 $$A^{0}$$

111$$A^{0}$$

211 $$A^{0}$$

311 $$A^{0}$$

Solution

Question 8

0

$$\infty $$

finite

data insufficient

Solution

Question 9

Electromagnetic waves

Mechanical waves

Either mechanical or electromagnetic waves

Neither mechanical nor electromagnetic waves

Solution

Question 10

Photo electric effect

Compton effect

Theory of relativity

None of the above

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.