Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

After absorbing a slowly moving neutron of Mass $$m_N$$ (momentum $$\approx 0$$) a nucleus of mass M breaks into two nuclei of masses $$m_1$$ and $$5m_1$$ $$(6m_1=M+m_N)$$ respectively. If the de Broglie wavelength of he nucleus with mass $$m_1$$ is $$\lambda$$, the de Broglie wavelength of the nucleus will be.

A
$$5\lambda$$

B
$$\lambda/5$$

C
$$\lambda$$

D
$$25\lambda$$

Solution

Since initial momentum of the system is $$P_i=0$$

Hence final momentum will also be $$P_f=0$$ from conservation of momentum

Now, let momentum of two nuclei produced be $$P_1$$ and $$P_2$$.

Then, both momentum will be opposite to each other and $$P_1=P_2$$ in magnitude

Now, $$\lambda=\dfrac{h}{mv}=\dfrac{h}{p}$$

Since,$$P_1=P-2\implies \lambda_1=\lambda_2=\lambda$$

Answer-(C)
Question 2
1 / -0

For which of the following wave functions, does the heisenberg uncertainty relation read $$\Delta x\Delta {p_x} = h/2.$$

A
harmonic oscillator ground state

B
hydrogen atom ground state

C
particle in a box ground state

D
free particle

Solution
Question 3
1 / -0

Particle having mass $$100 mg$$ is moving with a speed of $$200
m/s$$. The associated wavelength of the particle is $$( h = 6.626 \times
10^{-34} js)$$

A
$$1.2 \times 10^{-29} m$$

B
$$3.3 \times 10^{-32} m$$

C
$$3.3 \times 10^{-35} m$$

D
$$3.3 \times 10^{-38} m$$

Solution
Question 4
1 / -0

Assume that a molecule is moving with the root mean square speed at temperature 300 K. The de Broglie wavelength of nitrogen molecule is (Atomic mass of nitrogen = 14.0076 u, h = $$6.63 \, \times \, 10^{-27} \, J \, s, \, k_B \, = \, 1.38 \, \times \, 10^{-23} J \, K^{-1}, \, 1 \, u \, = \, 1.66 \,. \times \, 10^{-27} \, kg)$$

A
$$2.75 \, \times \, 10^{-11} \, m$$

B
$$2.75 \, \times \, 10^{-12} \, m$$

C
$$3.24 \, \times \, 10^{-11} \, m$$

D
$$3.24 \, \times \, 10^{-12} \, m$$

Solution

Mean Kinetic energy if a molecule:
$$ \dfrac{1}{2}\, mv_{rms}^{2} \, = \, \dfrac{3}{2}k_{B}T$$

Here, m = $$2 \times \, 14.0067 \, = \, 28.02 \, u$$
As $$\lambda \, = \, \dfrac{h}{mv_{rms}} \, = \dfrac{h}{\sqrt{3mk_{B}T}}$$

$$= \, \dfrac{6.63 \, \times 10^{-34}}{\sqrt{3\times \left ( 28.02 \times 1.66 \times 10^{-27} \right )} \times 1.38 \times 10^{-23}\times 300}$$

$$= \, 2.75 \times 10 ^{-11}m $$
Question 5
1 / -0

If the momentum of an electron is changed by $$ \Delta p, $$ then the de-Brogile wavelength associated with it changes by $$0.05 \% $$ . The initial momentum of electron will be :

A
$$ \dfrac { \Delta p}{200} $$

B
$$ \dfrac { \Delta p}{199} $$

C
$$ 199 \Delta p $$

D
$$ -2000 \Delta p $$

Solution

The initial momentum is given as,

$$\dfrac{{d\lambda }}{\lambda } = - \dfrac{{dP}}{{{P_i}}}$$

$$\dfrac{{0.05}}{{100}} = - \dfrac{{\Delta P}}{{{P_i}}}$$

$$\dfrac{{0.05}}{{100}} = - \dfrac{{\Delta P}}{{{P_i}}}$$

$${P_i} = - 2000\Delta P$$

The initial momentum of electron is $$ - 2000\Delta P$$.
Question 6
1 / -0

A surface irradiated with light of wavelength 480 nm gives out electrons with maximum velocity v m/s, the cut off wavelength being 600 nm. The same surface would release electrons with maximum velocity 2v m/s if it is irradiated by light of wavelength.

A
325 nm

B
360 nm

C
384 nm

D
300 nm

Solution
Question 7
1 / -0

A proton and an$$\alpha - particle$$ accelerated through same voltage.The ratio of their de-broglie wavelength will be :

A
$$1 : 2$$

B
$$2\sqrt 2 : 1$$

C
$$\sqrt2 : 1$$

D
$$2 : 1$$

Solution

$$\lambda _\alpha = \lambda_P$$
$$\dfrac{h}{\sqrt{2m_\alpha q_\alpha V}} = \dfrac{h}{\sqrt{2m_P q_P V}}$$

and we know
$$m_\alpha = 4m_P$$
$$2q_\alpha = q_P$$

$$\dfrac{\lambda_P}{\lambda_a} = \dfrac{1}{2}$$

Hence (A) option is correct
Question 8
1 / -0

An electron is moving with velocity $$6.6 \times 10^3 mls.$$ The DE-Imbroglio wavelength associated with electron is $$(mass of electron = 9 \times 10^{-31}kg,$$ Plank's Constant = $$6.62 \times 10^{-34} J-S)$$

A
$$1 \times 10^{-19} m$$

B
$$1 \times 10^{-5} m$$

C
$$1 \times 10^{-7} m$$

D
$$1 \times 10^{-10} m$$

Solution

de broglie wavelength, $$\lambda =\dfrac{h}{P}$$

$$\lambda = \dfrac{6.62 \times 10^{-34} }{g \times 10^{-31} \times 6.6 \times 10^3 m/s}$$

$$\lambda = 1 \times 10^{-7} m $$

Hence (C) option is correct
Question 9
1 / -0

The de-Broglie wavelength $$(\lambda_{B})$$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $$(\lambda_{G})$$ by

A
$$\lambda_{B} = \lambda_{G}$$

B
$$\lambda_{B} = 3\lambda_{G}$$

C
$$\lambda_{B} = 3\lambda_{\dfrac {G}{3}}$$

D
$$\lambda_{B} = 3\lambda_{\dfrac {G}{2}}$$

Solution
Question 10
1 / -0

The value of $$\left( { n }_{ 2 }+{ n }_{ 1 } \right) $$ and $$\left( { n }_{ 2 }^{ 2 }-{ n }_{ 1 }^{ 2 } \right) $$ for $${ He }^{ + }$$ ion in atomic spectrum are 4 and 8 respectively. The wavelength of emitted photon when electron jump from $${ n }_{ 2 }$$ to $${ n }_{ 1 }$$ is:

A
$$\dfrac { 32 }{ 9 } { R }_{ H }$$

B
$$\dfrac { 9 }{ 32 } { R }_{ H }$$

C
$$\dfrac { 9 }{ 32{ R }_{ H } } $$

D
$$\dfrac { 32 }{ 9{ R }_{ H } } $$

Solution

Given that
$$n_2+n_1=4$$ .....(1)
$$n_2^2-n^2_1=8$$ -----(2)
Here,
$$n_1$$ and $$n_2$$ are the energy levels of $$He^+$$ ion.

Solving equation (2), we have,

$$n_2^2 - n_1^2 = (n_2+n_1)(n_2-n_1)=8$$ -----(3)

Substituting (1) in equation (3), we have,

$$4\times(n_2-n_1) = 8$$
$$\Rightarrow (n_2-n_1) = \dfrac{8}{4}=2$$ ------- (4)

Solving equations (1) and (4), i.e.
$$n_2+n_1=4$$ .....(1)
$$n_2-n_1=2$$ .....(4)

We get,
$$n_1=1,n_2=3$$

Now, using Rydberg's wavelength equation, we have,

$$\dfrac{1}{\lambda}=R_HZ^2(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$$

where,
$$\lambda =$$ wavelength of the emitted photon
$$R_H =$$ Rydberg constant
$$Z =$$ atomic number of H or H-like ion and
$$n_1$$ & $$ n_2$$ are the energy levels of H or H-like ion

$$\Rightarrow \dfrac{1}{\lambda}=R_H \times 2^2\times (\dfrac{1}{1^2}-\dfrac{1}{3^2})$$

$$\Rightarrow \dfrac{1}{\lambda}=R_H \times 2^2\times (\dfrac{1}{1}-\dfrac{1}{9})$$

$$\Rightarrow \dfrac{1}{\lambda}=R_H \times 2^2\times (\dfrac{9-1}{9})$$

$$\Rightarrow \dfrac{1}{\lambda}=R_H\times 4\times \dfrac{8}{9} = \dfrac{32R_H}{9}$$

$$\Rightarrow \lambda=\dfrac{9}{32R_H}$$

$$\therefore$$ (C) option is correct.

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Dual Nature of Radiation and Matter Test - 69

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Dual Nature of Radiation and Matter Test - 69

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Question 1

$$5\lambda$$

$$\lambda/5$$

$$\lambda$$

$$25\lambda$$

Solution

Question 2

harmonic oscillator ground state

hydrogen atom ground state

particle in a box ground state

free particle

Solution

Question 3

$$1.2 \times 10^{-29} m$$

$$3.3 \times 10^{-32} m$$

$$3.3 \times 10^{-35} m$$

$$3.3 \times 10^{-38} m$$

Solution

Question 4

$$2.75 \, \times \, 10^{-11} \, m$$

$$2.75 \, \times \, 10^{-12} \, m$$

$$3.24 \, \times \, 10^{-11} \, m$$

$$3.24 \, \times \, 10^{-12} \, m$$

Solution

Question 5

$$ \dfrac { \Delta p}{200} $$

$$ \dfrac { \Delta p}{199} $$

$$ 199 \Delta p $$

$$ -2000 \Delta p $$

Solution

Question 6

325 nm

360 nm

384 nm

300 nm

Solution

Question 7

$$1 : 2$$

$$2\sqrt 2 : 1$$

$$\sqrt2 : 1$$

$$2 : 1$$

Solution

Question 8

$$1 \times 10^{-19} m$$

$$1 \times 10^{-5} m$$

$$1 \times 10^{-7} m$$

$$1 \times 10^{-10} m$$

Solution

Question 9

$$\lambda_{B} = \lambda_{G}$$

$$\lambda_{B} = 3\lambda_{G}$$

$$\lambda_{B} = 3\lambda_{\dfrac {G}{3}}$$

$$\lambda_{B} = 3\lambda_{\dfrac {G}{2}}$$

Solution

Question 10

$$\dfrac { 32 }{ 9 } { R }_{ H }$$

$$\dfrac { 9 }{ 32 } { R }_{ H }$$

$$\dfrac { 9 }{ 32{ R }_{ H } } $$

$$\dfrac { 32 }{ 9{ R }_{ H } } $$

Solution

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