Dual Nature of Radiation and Matter Test

Result

Dual Nature of Radiation and Matter Test - 39

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

If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is :

A
Zero

B
Less than that of a proton

C
More than that of a proton

D
Equal to that of a proton

Solution

de Broglie wavelength $$\lambda = \dfrac{h}{\sqrt{2mK}}$$

We get $$K = \dfrac{h^2}{2m \lambda^2}$$

$$\implies$$ $$K\propto \dfrac{1}{m}$$

We know that $$m_p > m_e$$

$$\therefore$$ $$\dfrac{K_e}{K_p} = \dfrac{m_p}{m_e} >1$$

$$\implies$$ $$K_e > K_p$$
Question 2
1 / -0

If alpha particle, proton and electron move with the same momentum, then their respective de-Broglie wavelengths $${ \lambda }_{ \alpha },{ \lambda }_{ p },{ \lambda }_{ e }$$ are related as

A
$${ \lambda }_{ \alpha }={ \lambda }_{ p }={ \lambda }_{ e }$$

B
$${ \lambda }_{ \alpha }<{ \lambda }_{ p }<{ \lambda }_{ e }$$

C
$${ \lambda }_{ \alpha }>{ \lambda }_{ p }>{ \lambda }_{ e }$$

D
$${ \lambda }_{ p }>{ \lambda }_{ e }>{ \lambda }_{ \alpha }$$

E
$${ \lambda }_{ p }<{ \lambda }_{ e }<{ \lambda }_{ \alpha }$$

Solution

Wavelength $$\lambda =\dfrac { h }{ p } $$
Since, momentum is same for all particles.
So, wavelength will be same for all
i.e., $${ \lambda }_{ \alpha }={ \lambda }_{ p }={ \lambda }_{ e }$$
Question 3
1 / -0

If alpha particle and deutron move with velocity $$v$$ and $$2v$$ respectively, the ratio of their de-Broglie wavelength will be _____

A
$$2 : 1$$

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

C
$$1 : 1$$

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

Solution

Key Concept According to de-Broglie wavelength, the wave associated with moving particle is given by
$$\lambda = \dfrac {h}{mv}$$
where, $$m$$ and $$v$$ are the mass and velocity of the particle and $$h$$ is Planck's constant.
de-Broglie wavelength of a particle, is given as
$$\lambda = \dfrac {h}{P}\Rightarrow \lambda \propto \dfrac {1}{P}$$
$$\Rightarrow \dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {P_{2}}{P_{1}}$$ or $$\dfrac {\lambda_{\alpha}}{\lambda_{D}} = \dfrac {P_{D}}{P_{1}}$$
$$\Rightarrow \dfrac {\lambda_{\alpha}}{\lambda_{D}} = \dfrac {2\times 2v}{4\times 2v} = \dfrac {4v}{4v}$$
$$\therefore \lambda_{\alpha} : \lambda_{D} = 1 : 1$$.
Question 4
1 / -0

If the kinetic energy of free electron is made double; the new de-Broglie wavelength will be __________ times that of initial wavelength.

A
$$\dfrac { 1 }{ \sqrt { 2 } } $$

B
$$\sqrt { 2 } $$

C
$$2$$

D
$$\dfrac { 1 }{ 2 } $$

Solution

We know that
de-Broglie wavelength $$\lambda =\dfrac { h }{ p } $$
$$\lambda =\dfrac { h }{ mv } $$
If $$K$$ be the kinetic energy of the electron, then
$$K=\dfrac { 1 }{ 2 } m{ v }^{ 2 }$$
$$v=\sqrt { { 2K }/{ m } } $$
So, de-Broglie wavelength is
$$\lambda =\dfrac { h }{ \sqrt { 2mK } } $$
$$\lambda \propto \dfrac { 1 }{ \sqrt { K } } $$
$$\therefore { \lambda }_{ 1 }=\dfrac { 1 }{ \sqrt { { K }_{ 1 } } } $$ ....(i)
$${ \lambda }_{ 2 }=\dfrac { 1 }{ \sqrt { 2{ K }_{ 1 } } } $$ ....(ii)
$$\dfrac { { \lambda }_{ 2 } }{ { \lambda }_{ 1 } } =\dfrac { { 1 }/{ \sqrt { 2{ K }_{ 1 } } } }{ { 1 }/{ \sqrt { { K }_{ 1 } } } } $$
$$\dfrac { { \lambda }_{ 2 } }{ { \lambda }_{ 1 } } =\dfrac { 1 }{ \sqrt { 2 } }$$
So, the new de-Broglie wavelength will be $$\dfrac { 1 }{ \sqrt { 2 } } $$ times that of initial wavelength.
Question 5
1 / -0

De-Broglie wavelength of a body of mass $$1$$kg moving with velocity of $$2000 \>$$m$$/$$s is?

A
$$3.32\times 10^{-27}\overset{o}{A}$$

B
$$1.5\times 10^7\overset{o}{A}$$

C
$$0.55\times 10^{-22}\overset{o}{A}$$

D
None of these

Solution

de Broglie wavelength $$\lambda \displaystyle =\frac{h}{mv}=\frac{6.6\times 10^{-34}}{1\times 2000}$$
$$=3.3\times 10^{-37}m=3.3\times 10^{-27}\overset{o}{A}$$.
Question 6
1 / -0

A body of mass $$100 g$$ moves at the speed of $$36 { km }/{ h }$$. The de-Broglie wavelength related to it is of the order _________ $$m$$.
$$\left( h=6.626\times { 10 }^{ -34 }Js \right) $$

A
$${ 10 }^{ -24 }$$

B
$${ 10 }^{ -14 }$$

C
$${ 10 }^{ -34 }$$

D
$${ 10 }^{ -44 }$$

Solution

Given, $$m=100 g=0.1 kg$$
$$v=36{ km }/{ h }=36\times \dfrac { 5 }{ 18 } { m }/{ s }=2\times 5=10{ m }/{ s }$$
$$h=6.626\times { 10 }^{ -34 }Js$$
We know that
$$\lambda =\dfrac { h }{ mv } $$ $$\left( \because p=\dfrac { h }{ \lambda } \quad and\quad p=mv \right) $$
$$=\dfrac { 6.626\times { 10 }^{ -34 } }{ 0.1\times 10 } $$
$$\lambda =6.626\times { 10 }^{ -34 }m$$
So, the de-Broglie wavelength related to it is of the order $${ 10 }^{ -34 }m$$
Question 7
1 / -0

In photoelectric effect the slope of straight line graph between stopping potential $$4(V_0)$$ and frequency of incident light (V) gives:

A
Charge on electron

B
Work function of emitter

C
Planek's Constant

D
ratio of Planek's constant to charge on electrons

Solution

We know that
By photoelectric equation we have
$$h\nu=h{\nu}_{o}+K.E$$
$$h\nu=W+e{V}_{o} \rightarrow$$shapping potential
$$e{V}_{o}=h\nu-W \Rightarrow {V}_{o}=\left( \cfrac{h}{e} \right)\nu+\left( \cfrac {-W}{e} \right)$$
$$y=mx+C$$
Compairing two equation $$\Rightarrow m=\cfrac {h}{c}$$
Slope is ratio of planck's constant to charge of electrons
Question 8
1 / -0

The de-Broglie wavelength of a neutron at $${ 27 }^{ o }C$$ is $$\lambda $$. What will be its wavelength at $${ 927 }^{ o }C$$?

A
$$\lambda /2$$

B
$$\lambda /3$$

C
$$\lambda /4$$

D
$$\lambda /9$$

Solution

$$\quad { \lambda }_{ neutron }\propto \cfrac { 1 }{ \sqrt { T } } \Rightarrow \cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\sqrt { \cfrac { { T }_{ 2 } }{ { T }_{ 1 } } } $$

$$\Rightarrow \cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\sqrt { \cfrac { (273+927) }{ (273+27) } } =2\Rightarrow { \lambda }_{ 2 }=\cfrac { \lambda }{ 2 } $$
Question 9
1 / -0

A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to

A
$$H$$

B
$$H^{1/2}$$

C
$$H^{0}$$

D
$$H^{-1/2}$$

Solution

Velocity acquired by a particle while talling from a height m is, $$\displaystyle v= \sqrt{2gH}$$
Thus, as, $$\displaystyle \lambda = \frac{h}{mv} = \frac{h}{m \sqrt{2gH}}$$
or $$\displaystyle \lambda \propto H^{-1/2}$$
Question 10
1 / -0

The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of that photon and the momentum of that electron is (c =velocity of light, h= Planck's constant)

A
$$h$$

B
$$c$$

C
$$\dfrac {1}{c}$$

D
None of these

Solution

We have $$\lambda_e=\dfrac {h}{mv}$$
and $$\lambda_p=\dfrac {h}{mc}$$
According to the question,
$$\lambda_e=\lambda_p$$
$$\therefore v=c$$
$$\dfrac {E_p}{P_e} =\dfrac {mc^2}{mv}= \dfrac {c^2}{c}= c$$

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 - 39

Result

Dual Nature of Radiation and Matter Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

Zero

Less than that of a proton

More than that of a proton

Equal to that of a proton

Solution

Question 2

$${ \lambda }_{ \alpha }={ \lambda }_{ p }={ \lambda }_{ e }$$

$${ \lambda }_{ \alpha }<{ \lambda }_{ p }<{ \lambda }_{ e }$$

$${ \lambda }_{ \alpha }>{ \lambda }_{ p }>{ \lambda }_{ e }$$

$${ \lambda }_{ p }>{ \lambda }_{ e }>{ \lambda }_{ \alpha }$$

$${ \lambda }_{ p }<{ \lambda }_{ e }<{ \lambda }_{ \alpha }$$

Solution

Question 3

$$2 : 1$$

$$1 : \sqrt {2}$$

$$1 : 1$$

$$\sqrt {2} : 1$$

Solution

Question 4

$$\dfrac { 1 }{ \sqrt { 2 } } $$

$$\sqrt { 2 } $$

$$2$$

$$\dfrac { 1 }{ 2 } $$

Solution

Question 5

$$3.32\times 10^{-27}\overset{o}{A}$$

$$1.5\times 10^7\overset{o}{A}$$

$$0.55\times 10^{-22}\overset{o}{A}$$

None of these

Solution

Question 6

$${ 10 }^{ -24 }$$

$${ 10 }^{ -14 }$$

$${ 10 }^{ -34 }$$

$${ 10 }^{ -44 }$$

Solution

Question 7

Charge on electron

Work function of emitter

Planek's Constant

ratio of Planek's constant to charge on electrons

Solution

Question 8

$$\lambda /2$$

$$\lambda /3$$

$$\lambda /4$$

$$\lambda /9$$

Solution

Question 9

$$H$$

$$H^{1/2}$$

$$H^{0}$$

$$H^{-1/2}$$

Solution

Question 10

$$h$$

$$c$$

$$\dfrac {1}{c}$$

None of these

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.