Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

The figure shows graph between stopping potential (along $$y$$-axis) and incident frequency (along x-axis) for different target metal $$A, B, C$$ and $$D$$. Let $$\displaystyle \phi _{A},\phi _{B},\phi _{C}\: and\: \phi _{D} $$are the work function of different target metals $$A, B, C$$ and $$D$$ respectively Then the correct rank of work function of different target metals is

A
$$\displaystyle \phi _{A}>\phi _{B}>\phi _{C}> \phi _{D} $$

B
$$\displaystyle \phi _{A}<\phi _{B}<\phi _{C}< \phi _{D} $$

C
$$\displaystyle \phi _{A}=\phi _{B}=\phi _{C}= \phi _{D} $$

D
can not be predict from the given data

Solution

The Einstein's photoelectric equation in terms of stopping potential is $$eV_s=h\nu-\phi$$
or $$V_s=\dfrac{h\nu}{e}-\phi/e$$
Thus, it will represent a straight line and intercept is $$\phi/e$$
From the graph we can write, $$\phi_A/e>\phi_B/e>\phi_C/e>\phi_D/e$$
or $$\phi_A>\phi_B>\phi_C>\phi_D$$
Question 2
1 / -0

The de-Broglie wavelength of an electron (mass $$1\times { 10 }^{ -30 }kg$$, charge$$=1.6\times { 10 }^{ -19 }\ C$$) with kinetic energy of $$200eV$$ is: (Planck's constant $$6.6\times { 10 }^{ -34 }J$$):

A
$$9.60\times { 10 }^{ -11 }m$$

B
$$8.25\times { 10 }^{ -11 }m$$

C
$$6.25\times { 10 }^{ -11 }m$$

D
$$5.00\times { 10 }^{ -11 }m$$

Solution

The de-Broglie wavelength $$\lambda =\cfrac { h }{ \sqrt { 2mk } } $$
Given $$h=6.6\times { 10 }^{ -34 }Js$$
$$m=1\times { 10 }^{ -30 }kg$$
$$K=200eV=200\times 1.6\times { 10 }^{ -19 }\quad J$$
Substituting all these values
$$\lambda =\cfrac { 6.6\times { 10 }^{ -34 } }{ \sqrt { 2\times 1\times { 10 }^{ -30 }\times 200\times 1.6\times { 10 }^{ -19 } } } $$
$$=0.825\times { 10 }^{ -10 }=8.25\times { 10 }^{ -11 }m$$
Question 3
1 / -0

An LED (Light Emitting Diode) is constructed from a p-n junction based on a certain $$Ga-As-P$$ semi-conducting material whose energy gap is $$1.9eV$$.What is the wavelength of the emitted light?

A
$$650nm$$

B
$$65\mathring { A } $$

C
$$800nm$$

D
$$8000\mathring { A } $$

Solution

The wavelength of emitted light
$$\lambda=\cfrac{hc}{{E}_{g}}$$
Where $${E}_{g}=$$ energy gap of semiconductor
$$=1.9eV$$
$$=1.9\times 1.6\times {10}^{-19}V$$
$$\lambda=\cfrac{6.6\times {10}^{-34}\times 3\times {10}^{8}}{1.9\times 1.6\times {10}^{-19}}m$$
$$6.5\times {10}^{-7}m$$
$$=650\times {10}^{-9}m$$
$$=650nm$$
Question 4
1 / -0

The force on a hemisphere of radius $$1$$ cm if a parallel beam of monochromatic light of wavelength $$500$$ nm. falls on it with an intensity of $$0.5$$ W/cm$$^{2}$$, striking the curved surface in a direction which is perpendicular to the flat face of the hemisphere is:
(Assume the collisions to be perfectly inelastic)

A
$$5.2\times 10^{-13} N$$

B
$$5.2\times 10^{-12} N$$

C
$$5.22\times 10^{-9} N$$

D
$$0$$ N

Solution

Momentum of each photon $$\Rightarrow p=\dfrac{h}{\lambda}=\dfrac{6.62\times10^{-34}}{500\times10^{-9}}=1.324\times10^{-27}kg.m/s$$

Number of photons per sq. cm $$\Rightarrow n=\dfrac{0.5\lambda}{hc}=\dfrac{0.5}{3\times10^8}\times\dfrac{1}{1.324\times10^{-27}}=1.258\times10^{18}photons/cm^2$$

As the beam is a parallel beam, photons strike on the circular area of the hemisphere $$\Rightarrow A=\pi r^2$$.

Force $$=$$ Number $$\times$$ Momentum$$ \times$$ Area $$=1.258\times1.324\times3.14\times1^2\times10^{-9}N$$

$$F=5.22\times10^{-9}N$$
Question 5
1 / -0

A photon of energy $$4\ eV$$ is incident on a metal surface whose work function is $$2\ eV$$. The minimum reverse potential to be applied for stopping the emission of electrons is :

A
$$2\ V$$

B
$$4\ V$$

C
$$6\ V$$

D
$$8\ V$$

Solution

When stopping potential is applied no electron will reach the cathode and the current will becomes zero.
Question 6
1 / -0

The de-Broglie wavelength of an electron moving with a velocity of $$1.5\times 10^8\ m/s$$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon $$(c=3\times 10^8\ m/s)$$:

A
$$2$$

B
$$4$$

C
$$\displaystyle\frac{1}{2}$$

D
$$\displaystyle\frac{1}{4}$$

Solution

Since the de-Broglie wavelength depends only on the momentum of particle, both electron and photon must have same momentum. Also, kinetic energy expression can be written as,

$$K=\dfrac { 1 }{ 2 } m{ v }^{ 2 }=\dfrac { 1 }{ 2 } pv$$

Hence,

$$\dfrac { { E }_{ electron } }{ { E }_{ photon } } =\dfrac { { v }_{ electron } }{ { v }_{ photon } } =\dfrac { 1.5 \times { 10 }^{ 8 } }{ 3 \times { 10 }^{ 8 } } =\dfrac { 1 }{ 2 } $$
Question 7
1 / -0

The de-Broglie wavelength of an electron is the same as that of a $$50 keV$$ X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $$0.5 MeV$$):

A
$$1 : 50$$

B
$$1 : 20$$

C
$$20 : 1$$

D
$$50 : 1$$

Solution

Energy of X-ray photon $$E_{photon} = 50 keV = 5\times 10^4 eV$$
Energy equivalent of electron mass is $$0.5MeV$$ i.e $$mc^2 = 0.5 MeV =5 \times 10^5 eV$$ ......(1)
Wavelength of the photon $$\lambda_{photon} = \dfrac{hc}{E} = \dfrac{1242}{5 \times 10^4}$$ $$ nm = 0.0248$$ $$ nm$$
$$\therefore$$ de-Broglie wavelength of the electron $$\lambda = \lambda_{photon} = 0.0248$$ $$nm$$

Kinetic energy of electron $$E_{electron} = \dfrac{h^2}{2m \times \lambda^2} = \dfrac{h^2c^2}{2 (mc^2) \lambda^2}$$ $$(\because E = \dfrac{p^2}{2m})$$

Kinetic energy of electron $$E_{electron} = \dfrac{(1242)^2}{2 (5\times 10^5) \times (0.0248)^2} $$ $$eV = 2.51 \times 10^3 eV$$
$$\therefore$$ $$\dfrac{E_{photon}}{E_{electron}} = \dfrac{5 \times 10^4}{2.51 \times 10^3} =20$$
Question 8
1 / -0

The ratio of the deBroglie wave length for the electron and proton moving with same velocity is:
[$$m_p$$- mass of propton, $$m_e$$-mass of electron]

A
$$m_p:m_e$$

B
$$m_p^2:m_e^2$$

C
$$m_e:m_p$$

D
$$m_e^2:m_p^2$$

Solution

$$\dfrac{\lambda_e}{\lambda_p}=\dfrac{\dfrac{h}{m_eV}}{\dfrac{h}{m_pV}}$$

$$=\dfrac{m_p}{m_e}$$
Question 9
1 / -0

The de-Broglie wavelength of a free electron with kinetic energy $$'E'$$ is $$\displaystyle \lambda $$. If the kinetic energy of the electron is doubled, the de-Broglie wavelength is:

A
$$\displaystyle \frac { \lambda }{ \sqrt { 2 } } $$

B
$$\displaystyle \sqrt { 2 } \lambda $$

C
$$\displaystyle \frac { \lambda }{ 2 } $$

D
$$\displaystyle 2\lambda $$

Solution

Kinetic energy initially is given by:
$$E=\dfrac{1}{2}mv^2=\dfrac{p^2}{2m}$$
$$\implies p=\sqrt {2m E}$$
where
$$m:$$ Mass of the particle
$$v:$$ Speed of the particle
$$p:$$ Momentum of the particle

De-broglie wavelength is given by,
$$\lambda=\dfrac{h}{p}$$
$$\therefore \lambda=\dfrac{h}{\sqrt{2mE}}$$

Given $$E_2=2E_1$$
$$\implies \lambda_2=\dfrac{\lambda_1} {\sqrt 2}$$
Given $$\lambda_1=\lambda, \quad \therefore \lambda_2=\dfrac{\lambda}{\sqrt 2}$$
Question 10
1 / -0

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition from an energy level with $$n = 4$$ to an energy level with $$n = 2?$$

A
$$5654A^o$$

B
$$3852A^o$$

C
$$4920A^o$$

D
$$4852A^o$$

Solution

Here 486 nm = 4860 $$A^o$$ (Approx. 4852 $$A^o$$)
Option D.

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Dual Nature of Radiation and Matter Test - 66

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Dual Nature of Radiation and Matter Test - 66

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Question 1

$$\displaystyle \phi _{A}>\phi _{B}>\phi _{C}> \phi _{D} $$

$$\displaystyle \phi _{A}<\phi _{B}<\phi _{C}< \phi _{D} $$

$$\displaystyle \phi _{A}=\phi _{B}=\phi _{C}= \phi _{D} $$

can not be predict from the given data

Solution

Question 2

$$9.60\times { 10 }^{ -11 }m$$

$$8.25\times { 10 }^{ -11 }m$$

$$6.25\times { 10 }^{ -11 }m$$

$$5.00\times { 10 }^{ -11 }m$$

Solution

Question 3

$$650nm$$

$$65\mathring { A } $$

$$800nm$$

$$8000\mathring { A } $$

Solution

Question 4

$$5.2\times 10^{-13} N$$

$$5.2\times 10^{-12} N$$

$$5.22\times 10^{-9} N$$

$$0$$ N

Solution

Question 5

$$2\ V$$

$$4\ V$$

$$6\ V$$

$$8\ V$$

Solution

Question 6

$$2$$

$$4$$

$$\displaystyle\frac{1}{2}$$

$$\displaystyle\frac{1}{4}$$

Solution

Question 7

$$1 : 50$$

$$1 : 20$$

$$20 : 1$$

$$50 : 1$$

Solution

Question 8

$$m_p:m_e$$

$$m_p^2:m_e^2$$

$$m_e:m_p$$

$$m_e^2:m_p^2$$

Solution

Question 9

$$\displaystyle \frac { \lambda }{ \sqrt { 2 } } $$

$$\displaystyle \sqrt { 2 } \lambda $$

$$\displaystyle \frac { \lambda }{ 2 } $$

$$\displaystyle 2\lambda $$

Solution

Question 10

$$5654A^o$$

$$3852A^o$$

$$4920A^o$$

$$4852A^o$$

Solution

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