Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

When the electron in a H-atom jumps from 6th excited state to 2nd excited state, the maximum number of emission lines will be.

A
$$20$$

B
$$18$$

C
$$10$$

D
$$5$$

Solution

Number lines = $$\begin{array}{l} \dfrac { { n\left( { n-1 } \right) } }{ 2 } \\ \dfrac { { 7\left( { 7-1 } \right) } }{ 2 } -\dfrac { { 3\left( { 3-1 } \right) } }{ 2 } =21-3=18 \end{array}$$
Question 2
1 / -0

Wavelength associated with an electron having kinetic energy $$ 3 \times 10^{-25} J$$ is $$x \times 10^{-7} \ m $$. What is the value of $$x$$?

A
9

B
4

C
5

D
6

Solution

$$\begin{array}{l} \lambda =\dfrac { h }{ { \sqrt { 2m\, K.E. } } } =x\times { 10^{ -7 } } \\ \Rightarrow x=\dfrac { { 6.63\times { { 10 }^{ -34 } }\times { { 10 }^{ 7 } } } }{ { \sqrt { 2\times 9\times { { 10 }^{ -31 } }\times 3\times { { 10 }^{ -25 } } } } } \\ =\dfrac { { 6.63\times { { 10 }^{ -27 } } } }{ { \sqrt { 54 } \times { { 10 }^{ -28 } } } } \\ =9.02. \\ Ans.\, (A) \end{array}$$
Question 3
1 / -0

Two particle of mass m and 2m moving in opposite direction collide head on. They have same de-Broglie wavelength before collision after the collision.

A
If e = 1, de-broglie wavelength of m is greater than that of 2 m

B
If e = 1, de-Broglie wavelength of 2 m is greater than that of m

C
The de-Broglie wavelength of m increases if e = 1

D
The de-Broglie wavelength of 2m remains same if e = 1
Question 4
1 / -0

Wavelength of an electron having energy 10 keV is ....................$${{\text{A}}^0}$$

A
0.12

B
1.2

C
12

D
120

Solution
Question 5
1 / -0

Among the following for which one mathemetical expression $$\lambda =\dfrac { h }{ p } $$ stands

A
De Broglie equation

B
Einstein

C
Uncertainty equation

D
Bohr equation

Solution
Question 6
1 / -0

A free particle with initial kinetic energy E and de-broglie wavelength $$\lambda$$ enters a region in which it has potential energy U. What is the particle's new de-Broglie wavelength?

A
$$\lambda ( 1 - U / E ) ^ { - 1 / 2 }$$

B
$$\lambda ( 1 - U / E )$$

C
$$\lambda ( 1 - U / E ) ^ { - 1 }$$

D
$$\lambda ( 1 - U / E ) ^ { 1 / 2 }$$

Solution

$$\begin{array}{l} The\, \, initial\, \, kinetic\, \, energy\, \, of\, \, free\, \, particle: \\ E=\frac { { { p^{ 2 } } } }{ { 2m } } \to =\sqrt { 2mE } \\ De\, \, broglie\, \, wavelength: \\ \lambda =\frac { h }{ p } =\frac { h }{ { \sqrt { 2mE } } } \\ Energy\, \, of\, \, the\, \, particle\, \, when\, \, it\, \, enters\, \, the\, \, region: \\ { E_{ f } }=E-U \\ So,\, \, its\, \, wavelength\, \, becomes: \\ { \lambda _{ f } }=\frac { h }{ { \sqrt { 2m{ E_{ f } } } } } =\frac { h }{ { \sqrt { 2m\left( { E-U } \right) } } } \\ \lambda _{ f }^{ 2 }=\frac { { { h^{ 2 } } } }{ { 2mE } } \left( { \frac { E }{ { E-V } } } \right) =\frac { { { h^{ 2 } } } }{ { 2mE } } \left( { \frac { 1 }{ { 1-U/E } } } \right) \\ { \lambda _{ f } }=\frac { h }{ { \sqrt { 2mE } } } { \left( { \frac { 1 }{ { 1-U/E } } } \right) ^{ 1/2 } }=\lambda { \left( { 1-U/E } \right) ^{ -1/2 } } \end{array}$$
Hence,
option $$(A)$$ is correct answer.
Question 7
1 / -0

Two particles of de-broglie wavelength $$\lambda_x$$ and $$\lambda_y$$ are moving in opposite directions. find dBroglie wavelength after perfectly inelastic collision:

A
$$\dfrac{\lambda_x \lambda_y}{\lambda_x - \lambda_y}$$

B
$$\dfrac{2\lambda_x \lambda_y}{\lambda_x - \lambda_y}$$

C
$$\dfrac{\lambda_x \lambda^2_y}{\lambda_x - \lambda_y}$$

D
$$\lambda_y - \lambda_x$$

Solution

$$\dfrac{h}{\lambda_x} - \dfrac{h}{\lambda_y} = \dfrac{h}{\lambda}$$

$$\dfrac{1}{\lambda} = \dfrac{\lambda_y - \lambda_x}{\lambda_x \lambda_y}$$

$$\dfrac{\lambda_x \lambda_y}{\lambda_y - \lambda_x}$$ or $$\dfrac{\lambda_x \lambda_y}{\lambda_x - \lambda_y}$$
Question 8
1 / -0

A proton and an $$\alpha$$ particle are accelerated through the same potential difference V. The ratio of their de Broglie wavelengths is?

A
$$\sqrt{2}$$

B
$$2\sqrt{2}$$

C
$$\sqrt{3}$$

D
$$2\sqrt{3}$$

Solution

$$\lambda =\dfrac{h}{\sqrt{2mqV}}\Rightarrow \lambda \sqrt{mq}=com$$
$$\Rightarrow \dfrac{\lambda_p}{\lambda_{\alpha}}=\sqrt{\dfrac{2\times 4}{1\times 1}}=2\sqrt{2}$$.
Question 9
1 / -0

A source of light has energy flux of 10 $$W/m^2$$. The light is falling perpendicularly to the surface of area 5 $$cm^2$$. If the surface completely absorbs the incident light then the force exerted on the surface is approximately

A
$$1.6 \times 10^{-11}$$ N

B
$$10^{-8}$$ N

C
$$3.2 \times 10^{-11}$$ N

D
$$2 \times 10^{-8}$$ N

Solution

$$\begin{array}{l}F=\frac{I A}{c} \ \ \ \text { (if coupletely absorbs) } \\F=\frac{10 \mathrm{~W} /\mathrm{m}^2\times 5 \times 10^{-4} \mathrm{~m}^{2}}{3 \times 10^{8}}\end{array}$$

$$\begin{aligned}F &=\frac{50}{3} \times 10^{-12}\mathrm{~N} \\F &=1.6 \times 10^{-11}\mathrm{~N} \\& \therefore \text { option } A\text { is correct }\end{aligned}$$
Question 10
1 / -0

The De-Broglie wave length of electron in second exited state of hydrogen atom is

A
$$5 {A}^\circ$$

B
$$10A^\circ$$

C
$$100A^\circ$$

D
$$6.6A^\circ$$

Solution

$$\lambda = \sqrt {\dfrac {150}{1.5}} = 10\overset {\circ}{A}$$.

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Re-Attempt Weekly Quiz Competition

Dual Nature of Radiation and Matter Test - 55

Result

Dual Nature of Radiation and Matter Test - 55

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SHARING IS CARING

Question 1

$$20$$

$$18$$

$$10$$

$$5$$

Solution

Question 2

9

4

5

6

Solution

Question 3

If e = 1, de-broglie wavelength of m is greater than that of 2 m

If e = 1, de-Broglie wavelength of 2 m is greater than that of m

The de-Broglie wavelength of m increases if e = 1

The de-Broglie wavelength of 2m remains same if e = 1

Question 4

0.12

1.2

12

120

Solution

Question 5

De Broglie equation

Einstein

Uncertainty equation

Bohr equation

Solution

Question 6

$$\lambda ( 1 - U / E ) ^ { - 1 / 2 }$$

$$\lambda ( 1 - U / E )$$

$$\lambda ( 1 - U / E ) ^ { - 1 }$$

$$\lambda ( 1 - U / E ) ^ { 1 / 2 }$$

Solution

Question 7

$$\dfrac{\lambda_x \lambda_y}{\lambda_x - \lambda_y}$$

$$\dfrac{2\lambda_x \lambda_y}{\lambda_x - \lambda_y}$$

$$\dfrac{\lambda_x \lambda^2_y}{\lambda_x - \lambda_y}$$

$$\lambda_y - \lambda_x$$

Solution

Question 8

$$\sqrt{2}$$

$$2\sqrt{2}$$

$$\sqrt{3}$$

$$2\sqrt{3}$$

Solution

Question 9

$$1.6 \times 10^{-11}$$ N

$$10^{-8}$$ N

$$3.2 \times 10^{-11}$$ N

$$2 \times 10^{-8}$$ N

Solution

Question 10

$$5 {A}^\circ$$

$$10A^\circ$$

$$100A^\circ$$

$$6.6A^\circ$$

Solution

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