Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 42

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Question 1
1 / -0

A proton, a neutron, an electron and an $$\alpha$$-particle have same energy. Then their de Broglie wavelengths compare as :

A
$$\lambda_p\, = \,\lambda_n\, > \,\lambda_e \,> \,\lambda_\alpha$$

B
$$\lambda_\alpha\, < \,\lambda_p\, = \,\lambda_n \,< \,\lambda_e $$

C
$$\lambda_e\, < \,\lambda_p\, = \,\lambda_n \,> \,\lambda_\alpha$$

D
$$\lambda_e\, = \,\lambda_p\, = \,\lambda_n \, = \,\lambda_\alpha$$

Solution

Kinetic energy of particle, $$K = \dfrac{1}{2}mv^2 \,\,\,\, or \,\,\,\,\, mv = \sqrt{2mK}$$

de Broglie wavelength, $$\lambda = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2mK}}$$

For the given value of $$K, \lambda \alpha \dfrac{1}{\sqrt{m}}$$

$$\therefore \lambda_p : \lambda_n : \lambda_e : \lambda_{\alpha} = \dfrac{1}{\sqrt{m_p}} : \dfrac{1}{\sqrt{m_n}}: \dfrac{1}{\sqrt{m_e}} :\dfrac{1}{\sqrt{m_\alpha}}$$

Since $$m_p = m_n, \,\, hence\,\, \lambda_p = \lambda_n$$

$$ As\ m_{\alpha} > m_p, \,\, therefore \,\, \lambda_{alpha} < \lambda_p$$

As $$m_e < m_n, therefore \lambda_n < \lambda_n$$

Hence $$ \lambda_{alpha} < \lambda_p = \lambda_n < \lambda_e$$
Question 2
1 / -0

Which of these particles having the same kinetic energy has the largest de Broglie wavelength?

A
Electron

B
Alpha particle

C
Proton

D
Neutron

Solution

As $$\lambda = \dfrac{h}{\sqrt{2mK}} \, so \, \lambda \propto \dfrac{1}{\sqrt{m}}$$

Out of the given particles m is least for electron, therefore electron has the largest value of de Broglie wavelength.
Question 3
1 / -0

When the velocity of an electron increases, its de Broglie wavelength:

A
increases

B
decreases

C
remains same

D
may increase or decrease

Solution

$$\lambda _D =\dfrac{h}{P} =\dfrac{h}{mv}$$

$$\lambda _D \propto \dfrac{1}{v}$$
velocity of an electron increases, its de Broglie wavelength decreases.
Option B
Question 4
1 / -0

If h is Plancks constant, the momentum of a photon of wavelength 0.01 $$A^o$$ is

A
$$10^{-2}h$$

B
h

C
$$10^2$$

D
$$10^{12}h$$

Solution

Momentum of photo, $$p = \dfrac{E}{c} = \dfrac{h\nu}{c}$$where E is the energy of a photon and c is the velocity of light.

$$\therefore$$ $$p = \dfrac{hc}{c\lambda}$$ $$\,\,\,\,$$ $$\left[\because \nu=\dfrac{c}{\lambda}\right]$$

$$p=\dfrac{h}{\lambda} = \dfrac{h}{0.01\times 10^{-10}} = 10^{12}h$$
Question 5
1 / -0

A particle of mass $$4m$$ at rest decays into two particles of masses $$m$$ and $$3m$$ having non-zero velocities. The ratio of the de Broglie wavelengths of the particles 1 and 2 is:

A
$$\dfrac{1}{2}$$

B
$$\dfrac{1}{4}$$

C
$$2$$

D
$$1$$

Solution

According to law of conservation of linear momentum, two particles will have equal and opposite momentum.
The de Broglie wavelength is given by

$$\lambda = \dfrac{h}{p} \, \, \, \, \therefore \dfrac{\lambda_1}{\lambda_2} = 1$$
Question 6
1 / -0

Find the wavelength of the incident light if the stopping potential is 0.6 V.

A
326 nm

B
454 nm

C
524 nm

D
232 nm

Solution

According to Einstein's photoelectric equation
$$eV_0 = h\upsilon - \phi_0 = \dfrac{hc}{\lambda} - \phi_0$$ $$\left (\because \upsilon = \dfrac{c}{\lambda} \right )$$

$$eV_0 = \dfrac{hc}{\lambda} - \phi_0$$

or $$\lambda = \dfrac{hc}{(eV_0 + \phi_0)} = \dfrac{1242eV nm}{(0.6 + 2.14)eV} = 454nm$$
Question 7
1 / -0

Relativistic corrections become necessary when the expression for the kinetic energy $$\dfrac{1}{2} mv^2$$, becomes comparable with $$mc^2$$, where m is the mass of the particle. At what de Broglie waxelength will relativistic corrections become important for an electron?

A
$$\lambda \, = \, l \, nm$$

B
$$\lambda \, = \, l0 \, nm$$

C
$$\lambda \, = \, l0^{-1} \, nm$$

D
$$\lambda \, = \, l0^{-4} \, nm$$

Solution

We know that the de Broglie's equation is written as:
$$p=\dfrac{h}{\lambda}$$
$$mv=\dfrac{h}{\lambda}$$
velocity of electrn, $$v = \dfrac{h}{(m\lambda)}$$

The value of plank's constant is $$h = 6.6\times 10^{-34}Js$$
and the mass of the electron is $$m = 9 \times 10^{-31}kg$$

$$(a) $$
For $$\lambda_1 = 10nm = 10^{-8}m$$

$$v_1 = \dfrac{6.6\times 10^{-34}}{(9\times 10^{-31})\times 10^{-8}} $$

$$= \dfrac{2.2}{3} \times 10^5 \approx 10^5m/s$$

$$(b)$$
For $$\lambda_2 = 10^{-1}nm = 10^{-10}m$$

$$v_2 = \dfrac{6.6\times 10^{-34}}{9\times 10^{-31} \times 10^{-10}}$$
$$ \approx 10^7 m/s$$

$$(c)$$
For $$\lambda_3 = 10^{-4}nm = 10^{-13}m$$

$$v_3 = \dfrac{6.6\times 10^{-34}}{9\times 10^{-31} \times 10^{-13}}$$
$$ \approx 10^{10} m/s$$

$$(d)$$
For $$\lambda_4 = 10^{-6}nm = 10^{-15}m$$

$$v_4 = \dfrac{6.6\times 10^{-34}}{9\times 10^{-31} \times 10^{-15}} $$
$$\approx 10^{12} m/s$$

Since $$v_3$$ and $$v_4$$ are greater than velocity of light . So, the relativistic correction can occur in these speeds of the electron.

Thus, option (C) and (d) are required wavelengths.
Question 8
1 / -0

Consider the four gases hydrogen, oxygen, nitrogen and helium at the same temperature. Arrange them in the increasing order of the de Broglie wavelengths of their molecules.

A
Hydrogen, helium, nitrogen, oxygen

B
Oxygen, nitrogen, hydrogen, helium

C
Oxygen, nitrogen, helium, hydrogen

D
Nitrogen, oxygen, helium, hydrogen

Solution

de Broglie wavelength of a gas molecule $$\lambda = \dfrac{h}{\sqrt{3mk_BT}}$$
where, T = Absolute temperature
kg = Boltzmann's constant
m = Mass of gas molecule
For the same temperature $$\lambda \propto \dfrac{1}{\sqrt{m}}$$
As $$M_0 > M_N > M_{He} > M_{H2}$$

$$\therefore \lambda_O < \lambda_N < \lambda_{He} < \lambda_{H2}$$
Question 9
1 / -0

Two particles $$A_1$$ and $$A_2$$ of masses $$m_1, \, m_2 \, (m_1 \, > \, m_2)$$ have the same de Broglie wavelength. Then

A
their momenta are the same.

B
their energies are the same.

C
momentum of $$A_1$$ is less than momentum of $$A_2$$

D
energy of $$A_1$$ is more than the energy of $$A_2$$

Solution

As $$ \lambda = \dfrac{h}{p} \, or \, p = \dfrac{h}{\lambda} \, or \, p \propto \dfrac{1}{\lambda}$$

$$\therefore \dfrac{p_1}{p_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{\lambda}{\lambda} = 1 \, or \, p_1 = p_2$$ ..............(1)

Also $$ E = \dfrac{1}{2} \dfrac{p^2}{m} = \dfrac{1}{2m} \dfrac{h^2}{\lambda^2} \, \, \, \, \left(\therefore p = \dfrac{h}{\lambda}\right)$$

or $$ E \propto \dfrac{1}{m} \, \, \therefore \dfrac{E_1}{E_2} = \dfrac{m_2}{m_1} \, < \, 1 \, or \, E_1 \, < \, E_2$$ ................(2)
From above equation: Option A
Question 10
1 / -0

The de Broglie wavelength associated with a ball of mass 150 g travelling at 30 m $$s^{-1}$$ is

A
$$1.47 \, \times \, 10^{-34}m$$

B
$$1.47 \, \times \, 10^{-16}m$$

C
$$1.47 \, \times \, 10^{-19}m$$

D
$$1.47 \, \times \, 10^{-31}m$$

Solution

Mass of the bail, m = 150 g = 0.15 kg.
speed of the ball, v = 30 $$m \, s^{-1}$$
Momentum, $$p = mv = 0.15 \, \times \, 30 \, = \, 4.5 kg \, m \, s^{-1}$$
de Broglie wavelength,
$$\lambda = \dfrac{h}{p} \, = \, \dfrac{6.63 \, \times \, 10^{-34}}{4.5} \, = \, 1.47 \, \times \, 10^{-34} m$$

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Dual Nature of Radiation and Matter Test - 42

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Dual Nature of Radiation and Matter Test - 42

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Question 1

$$\lambda_p\, = \,\lambda_n\, > \,\lambda_e \,> \,\lambda_\alpha$$

$$\lambda_\alpha\, < \,\lambda_p\, = \,\lambda_n \,< \,\lambda_e $$

$$\lambda_e\, < \,\lambda_p\, = \,\lambda_n \,> \,\lambda_\alpha$$

$$\lambda_e\, = \,\lambda_p\, = \,\lambda_n \, = \,\lambda_\alpha$$

Solution

Question 2

Electron

Alpha particle

Proton

Neutron

Solution

Question 3

increases

decreases

remains same

may increase or decrease

Solution

Question 4

$$10^{-2}h$$

h

$$10^2$$

$$10^{12}h$$

Solution

Question 5

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

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

$$2$$

$$1$$

Solution

Question 6

326 nm

454 nm

524 nm

232 nm

Solution

Question 7

$$\lambda \, = \, l \, nm$$

$$\lambda \, = \, l0 \, nm$$

$$\lambda \, = \, l0^{-1} \, nm$$

$$\lambda \, = \, l0^{-4} \, nm$$

Solution

Question 8

Hydrogen, helium, nitrogen, oxygen

Oxygen, nitrogen, hydrogen, helium

Oxygen, nitrogen, helium, hydrogen

Nitrogen, oxygen, helium, hydrogen

Solution

Question 9

their momenta are the same.

their energies are the same.

momentum of $$A_1$$ is less than momentum of $$A_2$$

energy of $$A_1$$ is more than the energy of $$A_2$$

Solution

Question 10

$$1.47 \, \times \, 10^{-34}m$$

$$1.47 \, \times \, 10^{-16}m$$

$$1.47 \, \times \, 10^{-19}m$$

$$1.47 \, \times \, 10^{-31}m$$

Solution

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