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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357841 Velocity of a proton is $\frac{c}{20}$ . Associated de broglie wavelength is

1

2.64 \times 10^{- 24} m m

2

2.64 \times 10^{- 24} c m

3

2.64 \times 10^{- 14}

A^{\circ}

4

2.64 \times 10^{- 14} m

Explanation:

$λ = \frac{h}{m v}$
$∵ v = \frac{c}{20} = \frac{3 \times 10^{8}}{20} = 1.5 \times 10^{7} m / s$
$h = 6.62 \times 10^{- 34} J s, m = 1.67 \times 10^{- 27} k g$
$∴ λ = \frac{6.62 \times 10^{- 34}}{1.6 \times 10^{- 27} \times 1.5 \times 10^{7}}$
$= 2.64 \times 10^{- 14} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357842 A material particle with a rest mass $m_{0}$ is moving with speed of light $c$ . The de-Broglie wavelength associated is given by

1

h / m_{0} c

2

m_{0} c / h

3 zero

4 infinite

Explanation:

$λ = \frac{h}{m v}$
where, $m = \frac{m_{0}}{\sqrt{1 - v^{2} / c^{2}}}$
As $v = c,$ then
$\Rightarrow m = \frac{m_{0}}{\sqrt{1 - \frac{c^{2}}{c^{2}}}} = \frac{m_{0}}{\sqrt{1 - 1}} = \frac{m_{0}}{0}$
$\Rightarrow m = \infty,$
Hence, $λ = \frac{h}{\infty} = 0$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357843 The ratio of the respective de-Broglie wavelength associated with electrons accelerated from rest with the voltages $100 V, 200 V$ and $300 V$ is

1

1 : 2 : 3

2

1 : 4 : 9

3

1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}}

4

1 : \frac{1}{2} : \frac{1}{3}

Explanation:

As we know, de-Broglie wavelength
$\begin{matrix} λ = \frac{h}{\sqrt{2 m e V}} or λ = \frac{1}{\sqrt{V}} \\ ∴ λ_{1} : λ_{2} : λ_{3} = \frac{1}{\sqrt{V_{1}}} : \frac{1}{\sqrt{V_{2}}} : \frac{1}{\sqrt{V_{3}}} \\ = \frac{1}{\sqrt{100}} : \frac{1}{\sqrt{200}} : \frac{1}{\sqrt{300}} \\ = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}} \end{matrix}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357844 A proton moving with one tenth of velocity of light has a certain de-Broglie wavelength of $λ$ . An alpha particle having certain kinetic energy has the same de- Broglie wavelength $λ$ . The ratio of kinetic energy of proton and that of alpha particle is

1

1 : 4

2

4 : 1

3

1 : 2

4

2 : 1

Explanation:

As $V = \frac{c}{10}$ , we can assume that the velocity of the proton is in non - relativistic range.
$K E = \frac{p^{2}}{2 m} = {(\frac{h}{λ})}^{2} \times \frac{1}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Ratio of kinetic energy of proton and that alpha particle is $\frac{K E_{p}}{K E_{α}} = \frac{m_{μ}}{m_{p}} = \frac{4 m_{p}}{m_{p}} = 4 : 1$

JEE - 2023

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357841 Velocity of a proton is $\frac{c}{20}$ . Associated de broglie wavelength is

1

2.64 \times 10^{- 24} m m

2

2.64 \times 10^{- 24} c m

3

2.64 \times 10^{- 14}

A^{\circ}

4

2.64 \times 10^{- 14} m

Explanation:

$λ = \frac{h}{m v}$
$∵ v = \frac{c}{20} = \frac{3 \times 10^{8}}{20} = 1.5 \times 10^{7} m / s$
$h = 6.62 \times 10^{- 34} J s, m = 1.67 \times 10^{- 27} k g$
$∴ λ = \frac{6.62 \times 10^{- 34}}{1.6 \times 10^{- 27} \times 1.5 \times 10^{7}}$
$= 2.64 \times 10^{- 14} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357842 A material particle with a rest mass $m_{0}$ is moving with speed of light $c$ . The de-Broglie wavelength associated is given by

1

h / m_{0} c

2

m_{0} c / h

3 zero

4 infinite

Explanation:

$λ = \frac{h}{m v}$
where, $m = \frac{m_{0}}{\sqrt{1 - v^{2} / c^{2}}}$
As $v = c,$ then
$\Rightarrow m = \frac{m_{0}}{\sqrt{1 - \frac{c^{2}}{c^{2}}}} = \frac{m_{0}}{\sqrt{1 - 1}} = \frac{m_{0}}{0}$
$\Rightarrow m = \infty,$
Hence, $λ = \frac{h}{\infty} = 0$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357843 The ratio of the respective de-Broglie wavelength associated with electrons accelerated from rest with the voltages $100 V, 200 V$ and $300 V$ is

1

1 : 2 : 3

2

1 : 4 : 9

3

1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}}

4

1 : \frac{1}{2} : \frac{1}{3}

Explanation:

As we know, de-Broglie wavelength
$\begin{matrix} λ = \frac{h}{\sqrt{2 m e V}} or λ = \frac{1}{\sqrt{V}} \\ ∴ λ_{1} : λ_{2} : λ_{3} = \frac{1}{\sqrt{V_{1}}} : \frac{1}{\sqrt{V_{2}}} : \frac{1}{\sqrt{V_{3}}} \\ = \frac{1}{\sqrt{100}} : \frac{1}{\sqrt{200}} : \frac{1}{\sqrt{300}} \\ = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}} \end{matrix}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357844 A proton moving with one tenth of velocity of light has a certain de-Broglie wavelength of $λ$ . An alpha particle having certain kinetic energy has the same de- Broglie wavelength $λ$ . The ratio of kinetic energy of proton and that of alpha particle is

1

1 : 4

2

4 : 1

3

1 : 2

4

2 : 1

Explanation:

As $V = \frac{c}{10}$ , we can assume that the velocity of the proton is in non - relativistic range.
$K E = \frac{p^{2}}{2 m} = {(\frac{h}{λ})}^{2} \times \frac{1}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Ratio of kinetic energy of proton and that alpha particle is $\frac{K E_{p}}{K E_{α}} = \frac{m_{μ}}{m_{p}} = \frac{4 m_{p}}{m_{p}} = 4 : 1$

JEE - 2023

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357841 Velocity of a proton is $\frac{c}{20}$ . Associated de broglie wavelength is

1

2.64 \times 10^{- 24} m m

2

2.64 \times 10^{- 24} c m

3

2.64 \times 10^{- 14}

A^{\circ}

4

2.64 \times 10^{- 14} m

Explanation:

$λ = \frac{h}{m v}$
$∵ v = \frac{c}{20} = \frac{3 \times 10^{8}}{20} = 1.5 \times 10^{7} m / s$
$h = 6.62 \times 10^{- 34} J s, m = 1.67 \times 10^{- 27} k g$
$∴ λ = \frac{6.62 \times 10^{- 34}}{1.6 \times 10^{- 27} \times 1.5 \times 10^{7}}$
$= 2.64 \times 10^{- 14} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357842 A material particle with a rest mass $m_{0}$ is moving with speed of light $c$ . The de-Broglie wavelength associated is given by

1

h / m_{0} c

2

m_{0} c / h

3 zero

4 infinite

Explanation:

$λ = \frac{h}{m v}$
where, $m = \frac{m_{0}}{\sqrt{1 - v^{2} / c^{2}}}$
As $v = c,$ then
$\Rightarrow m = \frac{m_{0}}{\sqrt{1 - \frac{c^{2}}{c^{2}}}} = \frac{m_{0}}{\sqrt{1 - 1}} = \frac{m_{0}}{0}$
$\Rightarrow m = \infty,$
Hence, $λ = \frac{h}{\infty} = 0$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357843 The ratio of the respective de-Broglie wavelength associated with electrons accelerated from rest with the voltages $100 V, 200 V$ and $300 V$ is

1

1 : 2 : 3

2

1 : 4 : 9

3

1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}}

4

1 : \frac{1}{2} : \frac{1}{3}

Explanation:

As we know, de-Broglie wavelength
$\begin{matrix} λ = \frac{h}{\sqrt{2 m e V}} or λ = \frac{1}{\sqrt{V}} \\ ∴ λ_{1} : λ_{2} : λ_{3} = \frac{1}{\sqrt{V_{1}}} : \frac{1}{\sqrt{V_{2}}} : \frac{1}{\sqrt{V_{3}}} \\ = \frac{1}{\sqrt{100}} : \frac{1}{\sqrt{200}} : \frac{1}{\sqrt{300}} \\ = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}} \end{matrix}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357844 A proton moving with one tenth of velocity of light has a certain de-Broglie wavelength of $λ$ . An alpha particle having certain kinetic energy has the same de- Broglie wavelength $λ$ . The ratio of kinetic energy of proton and that of alpha particle is

1

1 : 4

2

4 : 1

3

1 : 2

4

2 : 1

Explanation:

As $V = \frac{c}{10}$ , we can assume that the velocity of the proton is in non - relativistic range.
$K E = \frac{p^{2}}{2 m} = {(\frac{h}{λ})}^{2} \times \frac{1}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Ratio of kinetic energy of proton and that alpha particle is $\frac{K E_{p}}{K E_{α}} = \frac{m_{μ}}{m_{p}} = \frac{4 m_{p}}{m_{p}} = 4 : 1$

JEE - 2023

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357841 Velocity of a proton is $\frac{c}{20}$ . Associated de broglie wavelength is

1

2.64 \times 10^{- 24} m m

2

2.64 \times 10^{- 24} c m

3

2.64 \times 10^{- 14}

A^{\circ}

4

2.64 \times 10^{- 14} m

Explanation:

$λ = \frac{h}{m v}$
$∵ v = \frac{c}{20} = \frac{3 \times 10^{8}}{20} = 1.5 \times 10^{7} m / s$
$h = 6.62 \times 10^{- 34} J s, m = 1.67 \times 10^{- 27} k g$
$∴ λ = \frac{6.62 \times 10^{- 34}}{1.6 \times 10^{- 27} \times 1.5 \times 10^{7}}$
$= 2.64 \times 10^{- 14} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357842 A material particle with a rest mass $m_{0}$ is moving with speed of light $c$ . The de-Broglie wavelength associated is given by

1

h / m_{0} c

2

m_{0} c / h

3 zero

4 infinite

Explanation:

$λ = \frac{h}{m v}$
where, $m = \frac{m_{0}}{\sqrt{1 - v^{2} / c^{2}}}$
As $v = c,$ then
$\Rightarrow m = \frac{m_{0}}{\sqrt{1 - \frac{c^{2}}{c^{2}}}} = \frac{m_{0}}{\sqrt{1 - 1}} = \frac{m_{0}}{0}$
$\Rightarrow m = \infty,$
Hence, $λ = \frac{h}{\infty} = 0$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357843 The ratio of the respective de-Broglie wavelength associated with electrons accelerated from rest with the voltages $100 V, 200 V$ and $300 V$ is

1

1 : 2 : 3

2

1 : 4 : 9

3

1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}}

4

1 : \frac{1}{2} : \frac{1}{3}

Explanation:

As we know, de-Broglie wavelength
$\begin{matrix} λ = \frac{h}{\sqrt{2 m e V}} or λ = \frac{1}{\sqrt{V}} \\ ∴ λ_{1} : λ_{2} : λ_{3} = \frac{1}{\sqrt{V_{1}}} : \frac{1}{\sqrt{V_{2}}} : \frac{1}{\sqrt{V_{3}}} \\ = \frac{1}{\sqrt{100}} : \frac{1}{\sqrt{200}} : \frac{1}{\sqrt{300}} \\ = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}} \end{matrix}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357844 A proton moving with one tenth of velocity of light has a certain de-Broglie wavelength of $λ$ . An alpha particle having certain kinetic energy has the same de- Broglie wavelength $λ$ . The ratio of kinetic energy of proton and that of alpha particle is

1

1 : 4

2

4 : 1

3

1 : 2

4

2 : 1

Explanation:

As $V = \frac{c}{10}$ , we can assume that the velocity of the proton is in non - relativistic range.
$K E = \frac{p^{2}}{2 m} = {(\frac{h}{λ})}^{2} \times \frac{1}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Ratio of kinetic energy of proton and that alpha particle is $\frac{K E_{p}}{K E_{α}} = \frac{m_{μ}}{m_{p}} = \frac{4 m_{p}}{m_{p}} = 4 : 1$

JEE - 2023

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