Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 56

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Question 1
1 / -0

A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths $$'\lambda_x'$$ and $$'\lambda_y'$$ respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is?

A
$$\lambda_x+\lambda_y$$

B
$$\dfrac{\lambda_x\lambda_y}{\lambda_x+\lambda_y}$$

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

D
$$\lambda_x-\lambda_y$$

Solution

By momentum conservation
$$P_x-P_y=P_p$$
$$\dfrac{h}{\lambda_x}-\dfrac{h}{\lambda_y}=\dfrac{h}{\lambda_p}$$
$$\lambda_p=\dfrac{\lambda_x\lambda_y}{|\lambda_y-\lambda_x|}$$.
Question 2
1 / -0

A particle of mass M at rest decays into two particles of masses $$m_{1}$$ and $$m_{2}$$, having non-zero velocities. The ration of the de-Broglie wavelengths of the particles, $$I_{1}$$/$$I_{1}$$ is

A
$$m_{1}$$/$$m_{2}$$

B
$$m_{2}$$/$$m_{1}$$

C
1.0

D
$$\omega_{2}/\omega_{1}$$

Solution

$$\begin{aligned}\lambda &=\frac{h}{\sqrt{2 mk_{e}}}=\frac{h}{\sqrt{2 m\times\frac{1}{2}mv^{2}}}\\\therefore\lambda\propto \frac{1}{\sqrt{m_{2}}}&\Rightarrow \lambda \propto \frac{1}{m} \\ [&\left.\therefore\frac{\lambda_{1}}{\lambda_{2}}=\frac{m_{2}}{m_{1}}\right] \\\therefore & \text { option } B \text { is }\end{aligned}$$
Question 3
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If a proton and electron have same de Broglie wavelength, then

A
Kinetic energy of electron $$<$$ kinetic energy of proton

B
Kinetic energy of electron $$=$$ kinetic energy of proton

C
Momentum of electron $$=$$ momentum of proton

D
Momentum of electron $$<$$ momentum of proton

Solution

$$\begin{array}{l}\text { If proton and electron have same de-broglie } \\\text { wavelength then momentum of electron is } \\\text { equals to momentum of proton }\end{array}$$
Question 4
1 / -0

Two particle are moving perpendicular to each other with de-Broglie wave length $$\lambda_1$$ and $$\lambda_2$$. If they collide and stick together, then the de-Broglie wave length of system after collision is:

A
$$\lambda =\dfrac { \lambda_1\lambda _2 }{ \sqrt {\lambda ^2_1 +\lambda ^2_2 }} $$

B
$$\lambda =\dfrac { \lambda _{ 1 } }{ \sqrt { \lambda ^ 2_ 1+\lambda^2_2 } }$$

C
$$\lambda =\dfrac {\sqrt { \lambda ^{ 2 }_{ 1 }+\lambda ^{ 2 }_{ 2 } } }{ \lambda_2 } $$

D
$$\lambda =\dfrac { \lambda _ 1\lambda _ 2 }{ \sqrt {\lambda_1+\lambda_2} } $$

Solution

$$P\hat { i } +P\hat { j } =\vec { P } $$
$$\dfrac { h }{ \lambda _{ 1 } } i+\dfrac { h }{ \lambda _{ 2 } } j=\dfrac { h }{ \lambda } $$
$$\dfrac { 1 }{ \lambda } =\sqrt { \dfrac { 1 }{ \lambda ^{ 2 }_{ 1 } } +\dfrac { 1 }{ \lambda ^ 2_ 2 } } $$
$$\lambda =\dfrac { \lambda _{ 1 }\lambda _{ 2 } }{\sqrt{ { \lambda ^ 2_ 1 }+\lambda ^2_2 } }$$
Question 5
1 / -0

In a H-atom, an electron is in $$2^{nd}$$ excited state and its radius $$= 4.75 \mathring{A}$$. Calculate de-broglie wavelength of the electron.

A
$$\dfrac{8.5\pi}{4} \mathring{A}$$

B
$$0$$

C
$$\dfrac{19\pi}{3} \mathring{A}$$

D
$$\dfrac{9.5\pi}{3} \mathring{A}$$

Solution

$$\lambda = \dfrac{h}{mv} = \dfrac{h}{\dfrac{nh}{2\pi r}} = \dfrac{2\pi r}{n}$$ or

$$\lambda = \dfrac{2\pi}{3} \times 4.75 \mathring{A} = \dfrac{9.5 \pi}{3} \mathring{A}$$
Question 6
1 / -0

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $$4.65 \mathring{A}$$). The de-Broglie wavelength of this electron is:

A
$$12.9 \mathring{A}$$

B
$$3.5 \mathring{A}$$

C
$$9.7 \mathring{A}$$

D
$$6.6 \mathring{A}$$

Solution

$$2\pi r_n = n\lambda_n$$

$$\lambda_3 = \dfrac{2\pi(4.65 \times 10^{-10})}{3}$$

$$\lambda_3 = 9.7 {A}^\circ$$
Question 7
1 / -0

For the same speed, de Broglie wavelength.

A
Of electron is larger than proton

B
Of proton is larger than $$\alpha-particle$$

C
Of electron is larger than $$\alpha-particle$$

D
All of the above

Solution
Question 8
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Radiation of frequency, $$v$$, are incident on a photosensitive metal. The maximum kinetic energy of photoelectric is $$E$$. When the frequency of the incident radiation is doubled, what is the maximum kinetic energy of the photoelectrons?

A
$$2E$$

B
$$E/2$$

C
$$E+ hv$$

D
$$E-hv$$

Solution
Question 9
1 / -0

Two particles A and B of same mass have their total energies $$E_A$$ and $$E_B$$ in the ratio $$E_A:E_B=1:2$$. Their potential energies $$U_A$$ and $$U_B$$ are in the ratio $$U_A:U_B=1:2$$ . If $$\lambda_A$$ and $$\lambda_B$$ are their de-Broglie wavelengths, then $$\lambda_A:\lambda_B$$ is

A
$$1:2$$

B
$$2:1$$

C
$$1:\sqrt{2}$$

D
$$\sqrt{2}:1$$

E
$$1:1$$

Solution

$$\dfrac{E_A}{E_B}=\dfrac{1}{2}$$,
$$\dfrac{U_A}{U_B}=\dfrac{1}{2}$$
So $$E_A=x,E_B=2x$$
And $$U_A=y,U_B=2y$$
$$\because E_A=U_A+K_A$$
And $$E_B=U_B+K_B$$
here $$K_A$$ and $$K_B$$ are kinetic energy of particles A and B So
$$K_A=E_A-U_A=(x-y)$$.........(i)
$$K_B=E_B-U_B=2(x-y)$$.........(i)
$$\because$$ de-Broglie wavelength,
$$\lambda=\dfrac{h}{\sqrt{2mk}}$$

So, $$\lambda_A=\dfrac{h}{\sqrt{2mK_A}}$$=$$\lambda_B=\dfrac{h}{\sqrt{2mK_B}}$$

$$\therefore \dfrac{\lambda_A}{\lambda_B}=\sqrt{\dfrac{K_B}{K_A}}$$...........(iii)
From Equation (i), (ii) and (iii),

$$\dfrac{\lambda_A}{\lambda_B}=\sqrt{\dfrac{2(x-y)}{(x-y)}}=\dfrac{\sqrt{2}}{1}$$
Question 10
1 / -0

The de-Broglie wavelength of electron in $$3^{rd}$$ orbit of $$He^{+1}$$ ion is approximately.

A
$$2A^0$$

B
$$3A^0$$

C
$$4A^0$$

D
$$5A^0$$

Solution

We know that,

$$2\pi r=n\lambda$$

$$\lambda=\dfrac{2\pi r}{n}$$

$$\lambda=2\pi\times(0.529\ A^o)\times\dfrac{n^2}{z_{n}}$$

$$\lambda=2\pi\times(0.529\ A^o)\times\dfrac{n^2}{z}$$

$$\lambda=2\pi\times(0.529\ A^o)\dfrac{3}{2}$$

$$\lambda=3\pi\times(0.529\ A^o)≈5\ A^o$$

So, Te de broglie wavelength is $$5\ A^{o}$$

Hence, D is correct option.

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Re-Attempt Weekly Quiz Competition

Dual Nature of Radiation and Matter Test - 56

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Dual Nature of Radiation and Matter Test - 56

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Question 1

$$\lambda_x+\lambda_y$$

$$\dfrac{\lambda_x\lambda_y}{\lambda_x+\lambda_y}$$

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

$$\lambda_x-\lambda_y$$

Solution

Question 2

$$m_{1}$$/$$m_{2}$$

$$m_{2}$$/$$m_{1}$$

1.0

$$\omega_{2}/\omega_{1}$$

Solution

Question 3

Kinetic energy of electron $$<$$ kinetic energy of proton

Kinetic energy of electron $$=$$ kinetic energy of proton

Momentum of electron $$=$$ momentum of proton

Momentum of electron $$<$$ momentum of proton

Solution

Question 4

$$\lambda =\dfrac { \lambda_1\lambda _2 }{ \sqrt {\lambda ^2_1 +\lambda ^2_2 }} $$

$$\lambda =\dfrac { \lambda _{ 1 } }{ \sqrt { \lambda ^ 2_ 1+\lambda^2_2 } }$$

$$\lambda =\dfrac {\sqrt { \lambda ^{ 2 }_{ 1 }+\lambda ^{ 2 }_{ 2 } } }{ \lambda_2 } $$

$$\lambda =\dfrac { \lambda _ 1\lambda _ 2 }{ \sqrt {\lambda_1+\lambda_2} } $$

Solution

Question 5

$$\dfrac{8.5\pi}{4} \mathring{A}$$

$$0$$

$$\dfrac{19\pi}{3} \mathring{A}$$

$$\dfrac{9.5\pi}{3} \mathring{A}$$

Solution

Question 6

$$12.9 \mathring{A}$$

$$3.5 \mathring{A}$$

$$9.7 \mathring{A}$$

$$6.6 \mathring{A}$$

Solution

Question 7

Of electron is larger than proton

Of proton is larger than $$\alpha-particle$$

Of electron is larger than $$\alpha-particle$$

All of the above

Solution

Question 8

$$2E$$

$$E/2$$

$$E+ hv$$

$$E-hv$$

Solution

Question 9

$$1:2$$

$$2:1$$

$$1:\sqrt{2}$$

$$\sqrt{2}:1$$

$$1:1$$

Solution

Question 10

$$2A^0$$

$$3A^0$$

$$4A^0$$

$$5A^0$$

Solution

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