Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

If $$10,000V$$ are applied across an X-ray tube, find the ratio of wavelength of the incident electrons and the shortest wavelength of X-ray coming out of the X-ray tube, given $$e/m$$ of electron $$=1.8\times10^{11}\space C\space kg^{-1}$$.

A
$$1:10$$

B
$$10:1$$

C
$$5:1$$

D
$$1:5$$

Solution

$$V=10kV=10^4\ volts$$
The shortest wavelength is given as, $$\lambda_1=hc/eV_o$$
The de-broglie wavelength is given as $$\lambda_2=h/p=h/\sqrt{2\times eV_o\times m_e}$$ wher $$m_e$$ is the mass of the electron, $$p$$ is the momentum.
We can further write de-broglie wavelength as
$$\lambda_2=hc/(eV_o\sqrt{2V_oc^2/\dfrac{e}{m_e}})=hc/10eV_o$$
$$\Rightarrow \lambda_2=\lambda_1/10$$
So ratio of the de Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced is $$\lambda_2:\lambda_1=1:10$$
Question 2
1 / -0

The electron cannot exist inside the nucleus because

A
de-Broglie wavelength associated with electron in $$\beta$$ -decay is much less than the size of nucleus

B
de-Broglie wavelength associated with electron in $$\beta$$ -decay is much greater than the size of nucleus

C
de-Broglie wavelength associated with electron in $$\beta$$ -decay is equal to the size of nucleus

D
de-Broglie negative charge cannot exist in the nucleus

Solution

$${ \beta }^{ - }\quad decay$$:
$${ \beta }^{ - }$$ radioactivity occurs when a nucleus emits a negative electron from an unstable radioactive nucleus whose deBroglie wavelength of order $$MeV$$ is much larger than nuclear dimensions.
Question 3
1 / -0

The circumference of the second Bohr orbit of electron in hydrogen atom is $$600nm$$. The potential difference that must be applied between the plates so that the electrons have the de Broglie wavelength corresponding in this circumference is

A
$$10^{-5}\space V$$

B
$$\displaystyle\frac{5}{3}\times10^{-5}\space V$$

C
$$5\times10^{-5}\space V$$

D
$$3\times10^{-5}\space V$$

Solution

For second Bohr orbit, $$ 600nm=n\lambda =2\lambda \Rightarrow \lambda =3000{ A }^{ o }$$

For electron $$\lambda =\dfrac { 12.27 }{ \sqrt { V } } { A }^{ o }$$

Voltage $$ V={ \left( \dfrac { 12.27 }{ 3000 } \right) }^{ 2 }=\dfrac { 150.553 }{ 9 } \times { 10 }^{ -6 }=1.67\times { 10 }^{ -5 }=\dfrac { 5 }{ 3 } \times { 10 }^{ -5 }V$$
Question 4
1 / -0

On the basis of the uncertainty principle it can be proved that

A
Electrons exist inside the nucleus

B
Electrons do not exist inside the nucleus

C
Neutrons exist inside the nucleus

D
Protons exist inside the nucleus

Solution

The Heisenberg's uncertainty principle, The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time.
$$\Delta x\Delta p>h/2$$
or, $$\Delta E\Delta t>h/2$$
For electrons, the confinement energy (energy required to confine electron in nucleus) comes out to be very high as compared to actually observed. So electrons do not exist inside the nucleus.
Question 5
1 / -0

The de-Broglie wavelength of a proton $$(mass=1.6 \times 10 ^{-27} kg)$$ accelerated through a potential difference of 1 kV is :

A
$$600 \mathring {A}$$

B
$$0.9 \times 10^{-12}m$$

C
$$7 \mathring { A } $$

D
$$0.9 \times 10^{-19}nm$$

Solution

$$eV = KE = \cfrac{1}{2}mv^2$$

$$KE = \cfrac{p^2}{2m}$$

De-broglie wavelength,

$$ \lambda = \cfrac{h}{p}$$

$$ \lambda = \cfrac{h}{\sqrt{2mE}} = \cfrac{h}{\sqrt{2mVe}} $$

Putting the values of constants, we get:

$$ \lambda = \dfrac {6.6 \times 10^{-34}}{\sqrt {2 \times 1.6 \times 10^{-27} \times 1.6 \times 10^{-19} \times 1000 } }= 0.9 \times 10^{-12} m$$
Question 6
1 / -0

A particle with rest mass $$'m_0\ '$$ is moving with speed of light 'c'. The de-Broglie wavelength associated with it will be

A
$$\infty $$

B
Zero

C
$$m_0 \, c/h$$

D
$$hv/m_0c$$

Solution

Theory of special relativity, $$ m' = \gamma m = \cfrac{m}{\sqrt{1 -\cfrac{V^2}{c^2}}}$$

Debroglie wavelength, $$ \lambda = \cfrac{h}{m'v} = \cfrac{h \sqrt{1 -\cfrac{V^2}{c^2}}}{mv}$$
If $$v=c$$, then $$ \lambda =0$$
Question 7
1 / -0

If $$E_1, E_2, E_3$$ are the respective kinetic energies of an electron, an $$\alpha$$-particle and a proton, each having the same de-Broglie wavelength, then

A
$$E_1 > E_3 > E_2$$

B
$$E_2 > E_3 > E_1$$

C
$$E_1 > E_2 > E_3$$

D
$$E_1 = E_2 = E_3$$

Solution

$$KE = \cfrac{1}{2}mv^2$$
$$KE = \cfrac{p^2}{2m}$$
Debroglie wavelength
$$ \lambda = \cfrac{h}{p}$$
$$ \lambda = \cfrac{h}{\sqrt{2mE}}$$
Given
$$ \lambda_1 = \lambda_2 = \lambda_3$$
$$ m_1E_1 = m_2E_2 = m_3E_3$$
Since
$$m_1<m_3<m_2$$
$$E_1>E_3>E_2$$
Question 8
1 / -0

The de-Broglie wavelength of a neutron at $$972^o C$$ is $$\lambda$$.
What will be its wavelength at $$27^o C$$ ?

A
$$\dfrac{\lambda}{2}$$

B
$$\lambda$$

C
$$2\lambda$$

D
$$4\lambda$$

Solution

$$ E = \cfrac{1}{2}mv^2 = \cfrac{p^2}{2m}$$

De - Broglie wavelength

$$ \lambda = \cfrac{h}{p} = \cfrac{h}{\sqrt{2mE}}$$

Now for a molecule
$$ E = KT$$ (Boltzmann Law)

$$ \lambda = \cfrac{h}{\sqrt{2mkT}}$$

$$ \lambda = \cfrac{C}{\sqrt{T}}$$

$$ \cfrac{\lambda_1}{ \lambda_2} = \sqrt{\cfrac{T_2}{T_1}}$$

Here T is in K

$$ \cfrac{\lambda_1}{ \lambda_2} = \sqrt{\cfrac{300}{1245}}$$

$$ \lambda _2 = 2 \lambda_1$$
Question 9
1 / -0

A steel ball of mass m is moving with a kinetic energy 'K'. The de-Broglie wavelength associated with the ball is

A
$$\dfrac{h}{2mK}$$

B
$$\sqrt{\dfrac{h}{2 mk}}$$

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

D
Meaningless

Solution

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

$$p=mv$$

$$E = \cfrac{p^2}{2m}$$

$$ p = \sqrt{2mE}$$
De-Broglie wavelength

$$\lambda = \cfrac{h}{p} $$

$$ \lambda = \cfrac{h}{\sqrt{2mE}}$$
Question 10
1 / -0

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $$V_2-V_1=20 V$$. (i.e., plate 2 is at a higher potential). The plates are separated by $$d=0.1$$ m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $$1$$. What is its speed when it hits plate $$2$$?
$$(e=1.6 \times 10^{-19} C, m_e=9.11 \times 10^{-31} kg)$$

A
$$2.65 \times 10^6 m/s$$

B
$$7.02 \times 10^{12} m/s$$

C
$$1.87 \times 10^6 m/s$$

D
$$32 \times 10^{-19} m/s$$

Solution

The energy by accelerating the electron is in the form of kinetic energy.

So, $$ Ve = \cfrac{1}{2}mv^2$$

$$ v = \sqrt{\cfrac{2Ve}{m}}$$

$$ v =\sqrt{\cfrac{2 \times 20 \times (1.6 \times 10^{-19})}{9.1 \times 10^{-31}}}$$

$$v =2.65 \times 10^6 m/s$$

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Dual Nature of Radiation and Matter Test - 64

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Dual Nature of Radiation and Matter Test - 64

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SHARING IS CARING

Question 1

$$1:10$$

$$10:1$$

$$5:1$$

$$1:5$$

Solution

Question 2

de-Broglie wavelength associated with electron in $$\beta$$ -decay is much less than the size of nucleus

de-Broglie wavelength associated with electron in $$\beta$$ -decay is much greater than the size of nucleus

de-Broglie wavelength associated with electron in $$\beta$$ -decay is equal to the size of nucleus

de-Broglie negative charge cannot exist in the nucleus

Solution

Question 3

$$10^{-5}\space V$$

$$\displaystyle\frac{5}{3}\times10^{-5}\space V$$

$$5\times10^{-5}\space V$$

$$3\times10^{-5}\space V$$

Solution

Question 4

Electrons exist inside the nucleus

Electrons do not exist inside the nucleus

Neutrons exist inside the nucleus

Protons exist inside the nucleus

Solution

Question 5

$$600 \mathring {A}$$

$$0.9 \times 10^{-12}m$$

$$7 \mathring { A } $$

$$0.9 \times 10^{-19}nm$$

Solution

Question 6

$$\infty $$

Zero

$$m_0 \, c/h$$

$$hv/m_0c$$

Solution

Question 7

$$E_1 > E_3 > E_2$$

$$E_2 > E_3 > E_1$$

$$E_1 > E_2 > E_3$$

$$E_1 = E_2 = E_3$$

Solution

Question 8

$$\dfrac{\lambda}{2}$$

$$\lambda$$

$$2\lambda$$

$$4\lambda$$

Solution

Question 9

$$\dfrac{h}{2mK}$$

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

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

Meaningless

Solution

Question 10

$$2.65 \times 10^6 m/s$$

$$7.02 \times 10^{12} m/s$$

$$1.87 \times 10^6 m/s$$

$$32 \times 10^{-19} m/s$$

Solution

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