Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 24

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Question 1
1 / -0

A proton and an electron are accelerated by the same potential difference. Let $$\lambda_{e}$$ and $$\lambda_{p}$$ denote the de Broglie wavelength of the electron and the proton respectively, then

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

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

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

D
$$\lambda_{e} \geq \lambda _{p}$$

Solution

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

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

So de Broglie wavelength

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

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

So $$\lambda _{e}=\dfrac{h}{\sqrt{2m_{e}eV}}$$

$$\lambda _{p}=\dfrac{h}{\sqrt{2m_{p}eV}}$$

So, $$\lambda _{e}> \lambda _{p}$$
Question 2
1 / -0

An electron of mass $$9.1\times 10^{-31}$$kg and charge $$1.6\times 10^{-19}$$C is accelerated through a potential difference of 'V' volt. The de Broglie wavelength $$(\lambda )$$ associated with the electron is

A
$$\dfrac{12.27}{\sqrt{V}}A^{0}$$

B
$$\dfrac{12.27}{V}A^{0}$$

C
$$12.27\sqrt{V}A^{0}$$

D
$$\dfrac{1}{12.27\sqrt{V}}A^{0}$$

Solution

$$\dfrac{1}{2}mv^{2}=eV$$
$$mv=\sqrt{2meV}$$
Now $$\lambda =\dfrac{h}{mv}$$
$$=\dfrac{h}{\sqrt{2meV}}$$
$$=\dfrac{12.27}{\sqrt{V}}A^{0}$$
Question 3
1 / -0

The energy that should be added to an electron to reduce its de-Broglie wavelength from 1nm to 0.5nm is

A
Four times the intial energy

B
Equal to initial energy

C
Twice the initial energy

D
Thrice the intial energy

Solution

$$\dfrac{1}{2}mv^{2}=E$$
$$mv=\sqrt{2Em}$$
$$\lambda =\dfrac{h}{mv}$$
$$=\dfrac{h}{\sqrt{2Em}}$$
$$\lambda _{1}=\dfrac{h}{\sqrt{2E_{1}m}}$$
$$\lambda _{2}=\dfrac{h}{\sqrt{2E_{2}m}}$$
Given, $$\lambda_{1}=2\lambda_{2}$$
$$\dfrac{h}{\sqrt{2E_{1}m}}=2\dfrac{h}{\sqrt{2E_{2}m}}$$
$$\dfrac{1}{E_{1}}=\dfrac{4}{E_{2}}$$
$$E_{2}=4E_{1}$$
$$E_{2}=E_{1}+3E_{1}$$
Question 4
1 / -0

If an electron and a proton have the same kinetic energy, the ratio of the de Broglie wavelengths of proton and electron would approximately be :

A
1 : 1837

B
43 : 1

C
1837 : 1

D
1 : 43

Solution

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

$$p= mv= \sqrt{2mE}$$ $$= \sqrt{2mqv}$$

De Broglie wavelength associated with change particle

$$\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mE}} = \dfrac{h}{\sqrt{2mq \nu }}$$

where $$E = \dfrac{1}{2}mv^{2} = K.E, $$ P is momentum of change

So, $$\lambda _{P} = \dfrac{h}{\sqrt{2m_{p}E}}$$

$$\lambda _{e} = \dfrac{h}{\sqrt{2m_{e}E}}$$

So, $$\dfrac{\lambda _{p}}{\lambda _{e}} = \dfrac{\sqrt{m_{e}}}{\sqrt{m_{p}}}$$

$$= \sqrt {\dfrac{m_{e}}{m_{p}}}$$

$$= \sqrt {\dfrac{9.1 \times 10^{-31}}{1.67 \times 10^{-27}}}$$

$$= \sqrt{5.449 \times 10^{-4}}$$

$$= 2.33 \times 10^{-2}$$

$$= 0.0233$$

$$= \dfrac{1}{43}$$

So, the answer is option (D).
Question 5
1 / -0

A particle having a de Broglie wavelength of 1.0 $$A^{0}$$ is associated with a momentum of (given $$h = 6.6 \times 10^{-34}$$ Js)

A
$$6.6\times 10^{-26}kg$$ m/s

B
$$6.6\times 10^{-25}kg$$ m/s

C
$$6.6\times 10^{-24}kg$$ m/s

D
$$6.6\times 10^{-22}kg$$ m/s

Solution

Momentum of a particle with given de Broglie wavalength
$$=\dfrac{h}{\lambda }$$

$$=\dfrac{6.6\times 10^{-34}}{1\times 10^{-10}}$$

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

The de Broglie wavelength of a molecule of thermal energy KT (K is Boltzmann constant and T is absolute temperature) is given by :

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

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

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

D
$$\dfrac{1}{h\sqrt{2mKT}}$$

Solution

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

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

So de-broglie wavelength

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

$$\rightarrow\lambda=\dfrac{h}{\sqrt{2mKT}}$$
Question 7
1 / -0

The magnitude of the De-Broglie wavelength ($$\lambda$$) of an electron (e),proton(p),neutron (n) and $$\alpha $$ - particle ($$\alpha $$ ) all having the same energy of MeV, in the increasing order will follow the sequence:

A
$$\lambda_{e},\lambda _{p},\lambda_{n} ,\lambda_{\alpha } $$

B
$$\lambda_{\alpha },\lambda _{n},\lambda_{p} ,\lambda_{e}$$

C
$$\lambda_{e},\lambda _{n},\lambda_{p} ,\lambda_{\alpha }$$

D
$$\lambda_{p},\lambda _{e},\lambda_{\alpha } ,\lambda_{n }$$

Solution

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

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

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

$$\lambda _{e}=\dfrac{h}{\sqrt{2Em_{e}}}$$

$$\lambda _{p}=\dfrac{h}{\sqrt{2Em_{p}}}$$

$$\lambda _{n}=\dfrac{h}{\sqrt{2Em_{n}}}$$

$$\lambda _{\alpha }=\dfrac{h}{\sqrt{2Em_{\alpha }}}$$

we know $$m_{\alpha }> m_{n}> m_{p}> me$$
So $$\lambda _{\alpha }< \lambda _{n}< \lambda _{p}< \lambda e$$
Question 8
1 / -0

The de Broglie wavelength of an electron having 80 eV of energy is nearly
($$1eV=1.6\times10^{-19}J$$ , Mass of electron $$=9\times 10^{-31}kg$$,
Planck’s constant $$=6.6\times 10^{-34}$$Js) (nearly)

A
140 $$A^{0}$$

B
0.14 $$A^{0}$$

C
14 $$A^{0}$$

D
1.4$$A^{0}$$

Solution

$$given,\ \dfrac{1}{2}mv^{2}= 80\times 1.6\times 10^{-19}$$

$$mv=\ \sqrt{2\times 9\times 10^{-3.1}\times 80\times 1.6\times 10^{-19}}$$

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

$$=\dfrac{6.6\times 10^{34}}{\sqrt{2\times 9\times 10^{-31}\times 80\times 1.6\times 10^{-19}}}$$

$$=\ 1.4\ A^{0}$$
So, the answer is option (D).
Question 9
1 / -0

A particle of mass $$10^{-31}$$ kg is moving with a velocity equal to $$10^{5} ms^{-1}$$. The wavelength of the particle is equal to:

A
$$6.6\times 10^{-8}cm$$

B
$$0.66\times 10^{-8}cm$$

C
$$6.6\times 10^{-8}m$$

D
$$10cm$$

Solution

We know that
$$\lambda =\ \dfrac{h}{mv}$$
$$\lambda =\ \dfrac{6.63\times 10^{-34}}{10^{-31}\times 10^5}$$
$$\lambda =6.6\times 16^{-8}m$$
So, the answer is option (C).
Question 10
1 / -0

The de-Broglie wavelength of a particle moving with a velocity $$2.25 \times 10^{8}\ m/s$$ is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is :

A
$$\dfrac{1}{8}$$

B
$$\dfrac{3}{8}$$

C
$$\dfrac{5}{8}$$

D
$$\dfrac{7}{8}$$

Solution

$$K_{particle }=\dfrac 12 mv^2$$ also $$\lambda =\dfrac {h}{mv}$$
$$\Rightarrow K_{particle }=\dfrac 12 \left( \dfrac {h}{\lambda v}\right). v^2=\dfrac {vh}{2\lambda}$$ ...(i)
$$K_{photon}=\dfrac {hc}{\lambda}$$ ....(ii)
$$\therefore \dfrac {K_{particle }}{K_{photon}}=\dfrac {v}{2c}=\dfrac {2.25\times 10^8}{2\times 3\times 10^8}=\dfrac 38$$

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Dual Nature of Radiation and Matter Test - 24

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Dual Nature of Radiation and Matter Test - 24

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Question 1

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

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

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

$$\lambda_{e} \geq \lambda _{p}$$

Solution

Question 2

$$\dfrac{12.27}{\sqrt{V}}A^{0}$$

$$\dfrac{12.27}{V}A^{0}$$

$$12.27\sqrt{V}A^{0}$$

$$\dfrac{1}{12.27\sqrt{V}}A^{0}$$

Solution

Question 3

Four times the intial energy

Equal to initial energy

Twice the initial energy

Thrice the intial energy

Solution

Question 4

1 : 1837

43 : 1

1837 : 1

1 : 43

Solution

Question 5

$$6.6\times 10^{-26}kg$$ m/s

$$6.6\times 10^{-25}kg$$ m/s

$$6.6\times 10^{-24}kg$$ m/s

$$6.6\times 10^{-22}kg$$ m/s

Solution

Question 6

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

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

$$h\sqrt{2mKT}$$

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

Solution

Question 7

$$\lambda_{e},\lambda _{p},\lambda_{n} ,\lambda_{\alpha } $$

$$\lambda_{\alpha },\lambda _{n},\lambda_{p} ,\lambda_{e}$$

$$\lambda_{e},\lambda _{n},\lambda_{p} ,\lambda_{\alpha }$$

$$\lambda_{p},\lambda _{e},\lambda_{\alpha } ,\lambda_{n }$$

Solution

Question 8

140 $$A^{0}$$

0.14 $$A^{0}$$

14 $$A^{0}$$

1.4$$A^{0}$$

Solution

Question 9

$$6.6\times 10^{-8}cm$$

$$0.66\times 10^{-8}cm$$

$$6.6\times 10^{-8}m$$

$$10cm$$

Solution

Question 10

$$\dfrac{1}{8}$$

$$\dfrac{3}{8}$$

$$\dfrac{5}{8}$$

$$\dfrac{7}{8}$$

Solution

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