Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 25

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Weekly Quiz Competition

Question 1
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If $$E_{1},E_{2}$$ and $$E_{3}$$ are the kinetic energies of a proton, $$\alpha$$ -particle and deuteron respectively, which all have the same wavelength, then

A
$$E_{1}>E_{2}>E_{3}$$

B
$$E_{1}>E_{3}>E_{2}$$

C
$$E_{3}>E_{2}>E_{1}$$

D
$$E_{3}>E_{1}>E_{2}$$

Solution

The wavelength $$\displaystyle \lambda$$ is same for all three particles.
$$\displaystyle \lambda = \dfrac {h}{mu}$$
$$\displaystyle m$$ is mass and $$\displaystyle u$$ is velocity.
The product $$\displaystyle mu$$ will also be same.
The particle with higher m value will have lower u value
The kinetic energy $$\displaystyle = \dfrac {1}{2}mu^2 = \dfrac {1}{2} \times mu \times u$$
The particle with lower u value will have lower kinetic energy.
Thus, the particle with higher $$m$$ value will have lower kinetic energy.
The increasing order of mass is $$\displaystyle m_1 < m_3 < m_2$$.
The decreasing order of kinetic energy is $$\displaystyle E_1 > E_3 > E_2$$.
Note:, $$m_1$$, $$m_2$$ and $$m_3$$ are masses of proton, $$\displaystyle \alpha$$ particle and deuteron respectively.
Question 2
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If electron is having a wavelength of 100 $$A^{0}$$, then momentum is $$(gm \ cm \ s^{-1})$$ units

A
$$6.6\times 10^{-32}$$

B
$$6.6\times 10^{-29}$$

C
$$6.6\times 10^{-35}$$

D
$$6.6\times 10^{-21}$$

Solution

Momentum =$$\dfrac{h}{\lambda }$$

$$=\dfrac{6.6\times 10^{-34}}{100\times 10^{-10}}$$

$$=6.6\times 10^{-26}\ kg\ m/s$$
$$=6.6\times 10^{-26}\times 1000\times 100\ gm\ cm/s$$ ($$\because $$ 1kg=1000gm 1m=100cm)
$$=6.6\times 10^{-21}\ gm\ cm/s.$$
So, the answer is option (D).
Question 3
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The de-broglie wavelength of an electron and the wavelength of a photon are same. The ratio between the energy of the photon and the momentum of the electron is

A
h

B
c

C
1/h

D
1/c

Solution

De-broglie wavelength :
$$\lambda = \dfrac{h}{p}$$

$$\therefore \lambda_e= \dfrac{h}{m_{e}V}$$

and for proton

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

$$given\ \lambda _e = \lambda _p$$

$$\dfrac{h}{m_{e}V} = \dfrac{hc}{E}$$

$$\dfrac{E}{P_e}= c$$
So, the answer is option (B).
Question 4
1 / -0

The wavelength corresponding to a beam of electrons whose kinetic energy is 100 eV is
($$h=6.6\times 10^{-34}$$ Js, $$1eV=1.6\times10^{-19}J$$ J,$$m_{e}= 9.1 \times 10^{-31}$$ kg)

A
4.8 $$A^{0}$$

B
3.6 $$A^{0}$$

C
1.2 $$A^{0}$$

D
2.4 $$A^{0}$$

Solution

$$\dfrac{1}{2}mv^{2}=\ 100eV$$

$$mv=\sqrt{2\times 9.1\times 10^{-31}\times 100\times 1.6\times10^{-19}}$$

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

$$=\ 1.2\ A^{\circ}$$
So, the answer is option (C).
Question 5
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A proton and an alpha particle are accelerated through the same potential difference. The ratio of wavelengths associated with proton and alpha particle respectively is :

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

B
$$2:1$$

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

D
$$4:1$$

Solution

We know, $$\dfrac{1}{2}mv^2=eV$$

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

$$wavelength, \ \lambda = \dfrac{h}{mv}$$

$$\lambda _p = \dfrac{h}{\sqrt{2m_{p}q_{p}V}}$$

$$\lambda _p = \dfrac{h}{\sqrt{2meV}}$$

$$\lambda _{\alpha} = \dfrac{h}{\sqrt{2m_{\alpha} q_{\alpha}V}}$$

$$= \dfrac{h}{\sqrt{2\times4m\times2e\times V}}$$

$$\dfrac{\lambda _p}{\lambda _{\alpha }} = \dfrac{2\sqrt2}{1}$$

So, the answer is option (C).
Question 6
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A charged particle drops through V volts. Match the de Broglie wavelength for given particles
Particle $$\lambda\ in\ A^o$$
a. Electron e. $$\sqrt{\dfrac{0.0817}{V}}$$
b. Deuteron f. $$\sqrt{\dfrac{0.0102}{V}}$$
c. $$\alpha$$ particle g. $$\sqrt{\dfrac{150}{V}}$$
d. Proton h.$$\sqrt{\dfrac{0.0409}{V}}$$
.

A
a-g, b-e, c-h, d-f

B
a-g, b-h, c-f, d-e

C
a-h, d-e, c-f, d-g

D
a-h, h-f, c-e, d-h

Solution

$$\dfrac{1}{2}mv^{2}=qV$$

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

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

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

$$\lambda _{e}=\dfrac{h}{\sqrt{2m_e\times e\times V}}$$

$$\lambda _{e}=\sqrt{\dfrac{150}{V}}$$

$$\lambda _{\alpha }=\dfrac{h}{\sqrt{2\times 4\times m_p\times
2\times e\times V}}$$

$$\lambda _{\alpha }=\sqrt{\dfrac{0.0102}{V}}$$

$$\lambda _p=\sqrt{\dfrac{0.0817}{V}}$$

$$\lambda_d\ =\sqrt{\dfrac{0.0409}{V}}$$

$$\left ( m_d=2m_p\ \ q_{d}=e \right )$$
So, the answer is option (B).
Question 7
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If $$\lambda_{0} $$ is the de Broglie wavelength for a proton accelerated through a potential difference of 100V, the de Broglie wavelength for $$\alpha $$ -particle accelerated through the same potential difference is

A
$$2\sqrt{2}\lambda _{0}$$

B
$$\dfrac{\lambda _{0}}{2}$$

C
$$\dfrac{\lambda _{0}}{2\sqrt{2}}$$

D
$$\dfrac{\lambda _{0}}{\sqrt{2}}$$

Solution

Proton is accelerated through a potential difference 100 V
$$so,\ \dfrac{1}{2}mv^{2}= e\times 100$$

$$mv= \sqrt{2m_{p}e\times 100}$$

$$\lambda _o=\dfrac{h}{\sqrt{2me\times 100}}$$

$$De \ broglie \ wavelength\ of\ \alpha -\ particle$$

$$\lambda _{\alpha } = \dfrac{h}{\sqrt{2m_{\alpha }v_{\alpha }\times 100}}$$

$$\lambda _{\alpha } = \dfrac{h}{\sqrt{2\times 4m\times 2e\times 100}}$$

$$so\ \lambda _{\alpha} = \dfrac{\lambda _0}{2\sqrt{2}}$$
So, the answer is option (C).
Question 8
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If the energy of a particle is reduced to one fourth, then the percentage increase in its de Broglie wavelength will be :

A
$$4$$1%

B
$$141$$%

C
$$100$$%

D
$$71$$%

Solution

We know,

$$\dfrac{1}{2^{\cdot }}mv^{2}=E$$

$$mv=\sqrt{2mE}$$ &

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

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

$$\lambda ^{'}=\dfrac{h}{\sqrt{2m\dfrac{E}{4}}}$$

$$\dfrac{\lambda ^{'}}{\lambda }=2$$

$$\lambda ^{'}=2\lambda $$

So, the answer is option (C).
Question 9
1 / -0

Two particles of masses m and 2m have equal kinetic energies. Their de Broglie wavelengths are in the ratio of

A
$$1 :1$$

B
$$1:2$$

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

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

Solution

$$\dfrac{1}{2}mv^{2}=E$$

$$mv=\sqrt{2Em}$$
De-broglie wavelength

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

$$\lambda_m=\dfrac{h}{\sqrt{2E\times m}}$$

$$\lambda_{2m} =\dfrac{h}{\sqrt{2\times E\times 2m}}$$

$$\dfrac{\lambda_m}{\lambda_{2m}}=\dfrac{\sqrt{2}}{1}$$

So, the answer is option (D).
Question 10
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If the velocity of a particle is increased three times, then the percentage decrease in its de Broglie wavelength will be :

A
$$33.3$$%

B
$$66.6$$%

C
$$99.9$$%

D
$$22.2$$%

Solution

Wavelength, $$\lambda =\dfrac{h}{mv}$$

$$\lambda ^{'}=\dfrac{h}{3mv}$$

$$\dfrac{\lambda ^{'}}{\lambda }=\dfrac{1}{3}$$

$$\lambda ^{'}=\dfrac{\lambda }{3}$$

$$\Delta \lambda =\lambda -\dfrac{\lambda }{3}$$

$$=\dfrac{2\lambda }{3}$$

$$=66.6$$% decrease in$$\lambda$$

So, the answer is option (B).

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Re-Attempt Weekly Quiz Competition

a. Electron	e. $$\sqrt{\dfrac{0.0817}{V}}$$
b. Deuteron	f. $$\sqrt{\dfrac{0.0102}{V}}$$
c. $$\alpha$$ particle	g. $$\sqrt{\dfrac{150}{V}}$$
d. Proton	h.$$\sqrt{\dfrac{0.0409}{V}}$$

Dual Nature of Radiation and Matter Test - 25

Result

Dual Nature of Radiation and Matter Test - 25

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Question 1

$$E_{1}>E_{2}>E_{3}$$

$$E_{1}>E_{3}>E_{2}$$

$$E_{3}>E_{2}>E_{1}$$

$$E_{3}>E_{1}>E_{2}$$

Solution

Question 2

$$6.6\times 10^{-32}$$

$$6.6\times 10^{-29}$$

$$6.6\times 10^{-35}$$

$$6.6\times 10^{-21}$$

Solution

Question 3

h

c

1/h

1/c

Solution

Question 4

4.8 $$A^{0}$$

3.6 $$A^{0}$$

1.2 $$A^{0}$$

2.4 $$A^{0}$$

Solution

Question 5

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

$$2:1$$

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

$$4:1$$

Solution

Question 6

a-g, b-e, c-h, d-f

a-g, b-h, c-f, d-e

a-h, d-e, c-f, d-g

a-h, h-f, c-e, d-h

Solution

Question 7

$$2\sqrt{2}\lambda _{0}$$

$$\dfrac{\lambda _{0}}{2}$$

$$\dfrac{\lambda _{0}}{2\sqrt{2}}$$

$$\dfrac{\lambda _{0}}{\sqrt{2}}$$

Solution

Question 8

$$4$$1%

$$141$$%

$$100$$%

$$71$$%

Solution

Question 9

$$1 :1$$

$$1:2$$

$$1:\sqrt{2}$$

$$\sqrt{2}:1$$

Solution

Question 10

$$33.3$$%

$$66.6$$%

$$99.9$$%

$$22.2$$%

Solution

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