Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If the debroglie wave-length of $$He$$ at $$927^{o}C$$ is $$\lambda$$ then de-brolie wave-length of $$CH_{4}(g)$$ at $$27^{o}C$$ will be

A
$$\lambda$$

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

C
$$\dfrac {\lambda}{4}$$

D
$$2\lambda$$

Solution
Question 2
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If a light of wavelength $$\lambda$$ hits the moving electron, the uncertainty in measurement of its position will be

A
Greater than $$\lambda$$

B
Less than $$\lambda$$

C
Equal to $$\lambda$$

D
Any value

Solution
Question 3
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The ratio of de-Broglie wavelengths of molecules of hydrogen and helium Which are at temperature $$27^oC$$ and $$127^oC$$ respectively is

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

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

C
$$\sqrt { \dfrac { 8 }{ 3 } } $$

D
$$1$$

Solution

As we know,
de-Broglie wavelength $$\lambda =\dfrac{h}{mv_{rms}}, rms$$ velocity of a gas particle at the given temperature $$(T)$$ is given as

$$\dfrac 12 mv_{rms}^2=\dfrac 32 kT\Rightarrow v_{rms}=\sqrt{\dfrac{3\ kT}{m}}\Rightarrow mv_{rms}=\sqrt{3mkT}$$

$$\therefore \lambda =\dfrac{h}{mv_{rms}}=\dfrac{h}{\sqrt{3\ mkT}}$$

$$\Rightarrow \dfrac{\lambda_H}{\lambda_{He}}=\sqrt{\dfrac {m_{He} T_{He}}{m_H T_H}}=\sqrt{\dfrac{4(273+127)}{2(273+27)}}=\sqrt {\dfrac 83}$$
Question 4
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If $$E_{1},E_{2},E_{3}$$ and $$E_{4}$$ the respectively kinetic energies of electron, decuteron proton and neutron having same de-Broglic wavelength, identity the correct in which those value would increase,

A
$$E_{2},E_{4},E_{3},E_{1}$$

B
$$E_{1},E_{3},E_{4},E_{2}$$

C
$$E_{2},E_{4},E_{1},E_{3}$$

D
$$E_{3},E_{1},E_{2},E_{4}$$

Solution
Question 5
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The debroglie wavelength of photoelectrons is $$1\mathring {A}$$. Its accelerating potential is :

A
$$150\ V$$

B
$$15.3\ V$$

C
$$12.3\ V$$

D
$$13.6\ V$$

Solution
Question 6
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Using the Heisenberg uncertainty principle arrange the following particles in the order of increasing lowest energy possible.
$$(I)$$ an electron in $$H_2$$ molecule
$$(II)$$ a $$H$$ atom in a $$H_2$$ molecule
$$(III)$$ a proton in the carbon nucleus
$$(IV)$$ a $$H_2$$ molecule within a nanotube

A
$$( \mathrm { I } ) < ( \mathrm { III } ) < ( \mathrm { II } ) < ( \mathrm { IV } )$$

B
$$( \mathrm { IV } ) < ( \mathrm { II } ) < ( \mathrm { I } ) < ( \mathrm { III } )$$

C
$$( \mathrm { II } ) < ( \mathrm { IV } ) < ( \mathrm { III } ) < ( \mathrm { I } )$$

D
$$( \mathrm { IV } ) < ( \mathrm { I } ) < ( \mathrm { II } ) < ( \mathrm { II } )$$

Solution

We know that
nanotude $$\simeq nm \simeq 10^{-6}m$$
$$\Delta x\ 5$$ more $$\Rightarrow AP\simeq $$ minimum $$\Rightarrow P$$ is moles
$$\Rightarrow E=\dfrac{P^{v}}{2m}5$$ minimum
So $$(A)$$ and $$(cC)$$ are ruled one
$$H$$ atom is $$H_{2}$$ molecule is order of molecular size $$\Delta x_{1}\simeq $$ molecular size
electron in $$H_{2}$$ molecule
$$\Delta x_{2}=$$ atoms size
$$\Delta x_{1} > \Delta x_{2}$$
$$\Rightarrow E_{1} < E_{2}$$
Hence option $$(B)$$
Question 7
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Cut off potentials for a metal in photoelectric effect for light of wavelength $$\lambda_{1},\lambda_{2}$$ and $$\lambda_{3}$$ is found to be $$V_{1}, V_{2}$$ and $$V_{3}$$ are in Arithmetic progression and $$\lambda_{1},\lambda_{2}$$ and $$\lambda_{3}$$ will be:

A
$$Arithmetic\ progression$$

B
$$Geometric\ progression$$

C
$$Harmonic\ progression$$

D
$$None$$

Solution
Question 8
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A particle of mass m kg and charge q coulomb is accelerated from rest through V volt: then the de-Broglie wavelength $$\lambda$$ associated with it is given by

A
$$\lambda = \frac{h}{\sqrt{mV}}$$

B
$$\lambda = \frac{h}{\sqrt{2mq}}$$

C
$$\lambda = \frac{h}{\sqrt{2mqV}}$$

D
$$\lambda = \frac{h}{\sqrt{2mV}}$$

Solution
Question 9
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The wavelength $$\lambda_{e}$$ of an electron and $$\lambda_{p}$$ of a photon of same energy $$E$$ are related by

A
$$\lambda_{p} \propto \lambda_{e}^{2}$$

B
$$\lambda_{p} \propto \lambda_{e}$$

C
$$\lambda_{p} \propto \sqrt {\lambda_{e}}$$

D
$$\lambda_{p} \propto \dfrac {1}{\sqrt {\lambda_{e}}}$$

Solution

$${ \lambda }_{ p }\alpha { \lambda }_{ e }^{ 2 }$$
Wavelength of electron
$${ \lambda }_{ e }=\dfrac { h }{ \sqrt { 2mE } } $$ and proton $${ \lambda }_{ p }=\dfrac { { h }_{ c } }{ E } $$
$$\Rightarrow { \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2mE } $$ or $$E=\dfrac { { h }_{ c } }{ { \lambda }_{ p } } $$
$$\therefore$$ $${ \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2m\dfrac { { h }_{ c } }{ { \lambda }_{ p } } } $$
$$\Rightarrow { \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2m{ h }_{ c } } { \lambda }_{ p }$$
$${ \lambda }_{ e }^{ 2 }\alpha { \lambda }_{ p }$$
Question 10
1 / -0

An electron is moving in $${ 2 }^{ nd }$$ excited orbit of H-atom Radius of orbit in terms of de-Broglie wavelength $$\lambda $$ of electron can be given as

A
$$\cfrac { \lambda }{ \pi } $$

B
$$\cfrac { 2\lambda }{ \pi } $$

C
$$\cfrac { 3\lambda }{ 2\pi } $$

D
$$\cfrac { \lambda }{ 2\pi } $$

Solution

From Bohr's postulate,
$$mvr=\dfrac{nh}{2\pi}$$. . . . . .(1)
de-broglie wavelength, $$\lambda=\dfrac{h}{mv}$$
$$mv=\dfrac{h}{\lambda}$$. . . . . .(2)
Substitute equation (2) in equation (1), we get
$$r\dfrac{h}{\lambda}=\dfrac{nh}{2\pi}$$
$$r=\dfrac{n\lambda}{2\pi}$$
For 2nd excited state, $$n=3$$
Radius of orbit, $$r=\dfrac{3\lambda}{2\pi}$$
The correct option is C.