Dual Nature of Radiation and Matter Test

Result

Dual Nature of Radiation and Matter Test - 73

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Weekly Quiz Competition

Question 1
1 / -0

A particular of mass M at rest decays into two particles of masses $${ m }_{ 1 }$$ and $${ m }_{ 2 }$$ having non-zero velocities. The ratio of debroglie wavelength of paticle is

A
$${ m }_{ 1 }/{ m }_{ 2 }$$

B
$${ m }_{ 2 }/{ m }_{ 1 }$$

C
$$\sqrt { { m }_{ 2 }/{ m }_{ 1 } } $$

D
$$1$$

Solution

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Question 2
1 / -0

The de Broglie wavelength of an electron moving with a velocity $$1.5\times 10^8ms^{-1}$$ is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the energy of photon is?(apply non relativistic formula for electron)

A
$$2$$

B
$$4$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{1}{4}$$

Solution

$${ K }_{ particle }=\frac { 1 }{ 2 } m{ v }^{ 2 }\quad \lambda =\frac { h }{ mv } $$
$${ K }_{ particle }=\frac { 1 }{ 2 } \left( \frac { h }{ \lambda v } \right) .{ v }^{ 2 }=\frac { vh }{ 2\lambda } ........(1)$$
$${ K }_{ photon }=\frac { hc }{ \lambda } ........(2)$$
$$\frac { { K }_{ particle } }{ { K }_{ photon } } =\frac { v }{ 2c } =\frac { 1.5\times { 10 }^{ 8 } }{ 2\times 3\times { 10 }^{ 8 } } =\frac { 1 }{ 4 } $$
Question 3
1 / -0

An electron and a proton have the same De Broglie wavelength. Then the kinetic energy of the electron is :

A
Zero

B
Infinity

C
Equal kinetic energy of proton

D
Greater than the kinetic energy of proton

Solution

We know that,
$$E=hv=\dfrac { { h }_{ c } }{ \lambda } $$ is the fundamental relation.
$$E=cp$$
$$\Rightarrow cp=\dfrac { { h }_{ c } }{ \lambda } $$ or $$\lambda =\dfrac { h }{ p } $$
Thus, it is evident that same de-Broglie wavelength implies, Total energies of the particles are equal.
Question 4
1 / -0

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $$\lambda _ { n } , \lambda _ { g }$$ be the de Broglie wavelength of the electron in the $$n ^ { t h }$$ state and the ground state respectively. Let $$\Lambda _ { n }$$ be the wavelength of the emitted photon in the transition from the $$n ^ { t h }$$ state to the ground state. For large n, (A, B are constants)

A
$$\Lambda _ { n } \approx A + \frac { B } { \lambda _ { n } ^ { 2 } }$$

B
$$\Lambda _ { n } \approx A + B \lambda _ { n }$$

C
$$\Lambda _ { n } ^ { 2 } \approx A + B \lambda _ { n } ^ { 2 }$$

D
$$\Lambda _ { n } ^ { 2 } \approx \lambda$$

Solution
Question 5
1 / -0

Two different photons of energies, 1 eV and 2.5 eV, fall on two identical metal plates having work function 0.5 eV, Then the ratio of maximum KE of the electrons emitted from the two surface is -

A
1 : 2

B
1 : 4

C
2 : 1

D
4 : 1

Solution
Question 6
1 / -0

A small potassium foil is placed perpendicular to the incident light at a distance of 0.5 m from a light source of 1.0 W. work function for the potassium foil is 1.8 eV. the radius of a potassium atom is $$ 1.3 \times 10^{-10} m $$ assuming wave nature of lights, time elapsed before the first electron is emitted is

A
17 h

B
17 ns

C
17 min

D
17 s

Solution
Question 7
1 / -0

At temperature $$T,$$ the average kinctic energy of any particle is $$\dfrac { 3 } { 2 } \mathrm { kT }.$$ The de Broglie wavelength follows the order:

A
Visible photon $$>$$ Thermal electron $$>$$ Thermal neutron

B
Thermal proton $$>$$ Thermal electron $$>$$ Visible photon

C
Visible photon $$>$$ Thermal neutron $$>$$ Thermal electron

D
Thermal proton $$>$$ Visible photon $$>$$ Thermal electron

Solution
Question 8
1 / -0

A radiation of wavelength $$\lambda$$ illuminates a metal and ejects photoelectrons of maximum kinetic energy of 1eV. Another radiation of wavelength $$\dfrac{\lambda}{3}$$ ejects photoelectrons of maximum kinetic energy of 4eV. What will be the work function of the metal?

A
1eV

B
0.5eV

C
2eV

D
3eV

Solution
Question 9
1 / -0

Energy of $$1\ mole$$ of radio wave photons with frequency of $$909\ KHz$$ is :-

A
$$6.02\times 10^{-28}J$$

B
$$3.62 \times 10^{-4}J$$

C
$$1.00\times 10^{-4}J$$

D
$$6.02\times 10^{-3}J$$

Solution
Question 10
1 / -0

The wave number of the series limiting line for the Lyman series for hydrogen atom is $$\left( {{\text{R}} = {\text{109678}}\,{\text{c}}{{\text{m}}^{ - 1}}} \right)$$

A
82259 $${{\text{c}}{{\text{m}}^{ - 1}}}$$

B
109678 $${{\text{c}}{{\text{m}}^{ - 1}}}$$

C
$$\,1.2157 \times {10^{ - 5}}$$$${{\text{c}}{{\text{m}}^{ }}}$$

D
$$\,9.1176 \times {10^{ - 6}}$$$${{\text{c}}{{\text{m}}^{ }}}$$

Solution