QBManager

Dual nature of radiation and Matter

142454 A particle of mass $m$ is projected from ground with velocity $u$ making angle $θ$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.

1

\infty

2

h / mu \sin θ

3

h / mu \cos θ

4

h / mu

Explanation:

B As we know that
de-Broglie wavelength $(λ) = \frac{h}{mv}$
At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin θ$
$∴$ the wavelength is given by, $λ = \frac{h}{mu \sin θ}$

AIIMS-26.05.2018(E)

Dual nature of radiation and Matter

142464 Which of the following figure represents the variation of particle momentum and associated de Broglie wavelength?

1

2

3

4

Explanation:

D According to the de-Broglie hypothesis wavelength, $λ = \frac{h}{p}$
$∴ λ \propto \frac{1}{p}$
From the above relegation we have rectangular hyperbola.

Hence, curve (d) is the correct option.

Karnataka CET-2006

Dual nature of radiation and Matter

142472 In Davission- Germer experiment maximum intensity is observed at

1

50^{\circ}

and

54 V

2

54^{\circ}

and

50 V

3

50^{\circ}

and

50 V

4

65^{\circ}

and

50 V

Explanation:

A The Davission - Germer experiment proves that an electron has dual nature i.e. it behaves as both wave and a particle. It shows that an electron beam could undergo diffraction when passed through the atomic crystal of nickel. In Davission - Germer experiments maximum intensity is observed at $50^{\circ}$ and $54^{\circ} V$ .
According to Bragg's law -
$n λ = 2 d \sin θ$
In Davission- Germer experiment maximum intensity is observed at $θ = 50^{\circ}$ and $V = 54 V$

MP PET-2008

Dual nature of radiation and Matter

142475 An electron of mass ' $m$ ', when accelerated through a potential $v$ has de-Broglie wavelength $λ$ . The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

λ (\frac{M}{m})

4

λ (\frac{m}{M})

Explanation:

B We know that,
de-Broglie wavelength, $λ = \frac{h}{\sqrt{2 mE}}$
$λ \propto \frac{1}{\sqrt{m}}$
$λ^{'} = λ \sqrt{\frac{m}{M}}$
Where, $M$ is the mass of the accelerated proton.

BITSAT-2006

Dual nature of radiation and Matter

142483 The minimum wavelength of $X$ - ray emitted from $X$ -ray machine operating at an accelerating potential of $V$ volts is:

1

\frac{hc}{eV}

2

\frac{Vc}{eh}

3

\frac{eh}{Vc}

4

\frac{eV}{hc}

Explanation:

A We know that,
$E_{max} = h v_{max} = eV$
$\frac{hc}{λ_{min}} = eV$
$λ_{min} = \frac{hc}{eV}$

UPSEE - 2005

Dual nature of radiation and Matter

142454 A particle of mass $m$ is projected from ground with velocity $u$ making angle $θ$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.

1

\infty

2

h / mu \sin θ

3

h / mu \cos θ

4

h / mu

Explanation:

B As we know that
de-Broglie wavelength $(λ) = \frac{h}{mv}$
At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin θ$
$∴$ the wavelength is given by, $λ = \frac{h}{mu \sin θ}$

AIIMS-26.05.2018(E)

Dual nature of radiation and Matter

142464 Which of the following figure represents the variation of particle momentum and associated de Broglie wavelength?

1

2

3

4

Explanation:

D According to the de-Broglie hypothesis wavelength, $λ = \frac{h}{p}$
$∴ λ \propto \frac{1}{p}$
From the above relegation we have rectangular hyperbola.

Hence, curve (d) is the correct option.

Karnataka CET-2006

Dual nature of radiation and Matter

142472 In Davission- Germer experiment maximum intensity is observed at

1

50^{\circ}

and

54 V

2

54^{\circ}

and

50 V

3

50^{\circ}

and

50 V

4

65^{\circ}

and

50 V

Explanation:

A The Davission - Germer experiment proves that an electron has dual nature i.e. it behaves as both wave and a particle. It shows that an electron beam could undergo diffraction when passed through the atomic crystal of nickel. In Davission - Germer experiments maximum intensity is observed at $50^{\circ}$ and $54^{\circ} V$ .
According to Bragg's law -
$n λ = 2 d \sin θ$
In Davission- Germer experiment maximum intensity is observed at $θ = 50^{\circ}$ and $V = 54 V$

MP PET-2008

Dual nature of radiation and Matter

142475 An electron of mass ' $m$ ', when accelerated through a potential $v$ has de-Broglie wavelength $λ$ . The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

λ (\frac{M}{m})

4

λ (\frac{m}{M})

Explanation:

B We know that,
de-Broglie wavelength, $λ = \frac{h}{\sqrt{2 mE}}$
$λ \propto \frac{1}{\sqrt{m}}$
$λ^{'} = λ \sqrt{\frac{m}{M}}$
Where, $M$ is the mass of the accelerated proton.

BITSAT-2006

Dual nature of radiation and Matter

142483 The minimum wavelength of $X$ - ray emitted from $X$ -ray machine operating at an accelerating potential of $V$ volts is:

1

\frac{hc}{eV}

2

\frac{Vc}{eh}

3

\frac{eh}{Vc}

4

\frac{eV}{hc}

Explanation:

A We know that,
$E_{max} = h v_{max} = eV$
$\frac{hc}{λ_{min}} = eV$
$λ_{min} = \frac{hc}{eV}$

UPSEE - 2005

Dual nature of radiation and Matter

142454 A particle of mass $m$ is projected from ground with velocity $u$ making angle $θ$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.

1

\infty

2

h / mu \sin θ

3

h / mu \cos θ

4

h / mu

Explanation:

B As we know that
de-Broglie wavelength $(λ) = \frac{h}{mv}$
At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin θ$
$∴$ the wavelength is given by, $λ = \frac{h}{mu \sin θ}$

AIIMS-26.05.2018(E)

Dual nature of radiation and Matter

142464 Which of the following figure represents the variation of particle momentum and associated de Broglie wavelength?

1

2

3

4

Explanation:

D According to the de-Broglie hypothesis wavelength, $λ = \frac{h}{p}$
$∴ λ \propto \frac{1}{p}$
From the above relegation we have rectangular hyperbola.

Hence, curve (d) is the correct option.

Karnataka CET-2006

Dual nature of radiation and Matter

142472 In Davission- Germer experiment maximum intensity is observed at

1

50^{\circ}

and

54 V

2

54^{\circ}

and

50 V

3

50^{\circ}

and

50 V

4

65^{\circ}

and

50 V

Explanation:

A The Davission - Germer experiment proves that an electron has dual nature i.e. it behaves as both wave and a particle. It shows that an electron beam could undergo diffraction when passed through the atomic crystal of nickel. In Davission - Germer experiments maximum intensity is observed at $50^{\circ}$ and $54^{\circ} V$ .
According to Bragg's law -
$n λ = 2 d \sin θ$
In Davission- Germer experiment maximum intensity is observed at $θ = 50^{\circ}$ and $V = 54 V$

MP PET-2008

Dual nature of radiation and Matter

142475 An electron of mass ' $m$ ', when accelerated through a potential $v$ has de-Broglie wavelength $λ$ . The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

λ (\frac{M}{m})

4

λ (\frac{m}{M})

Explanation:

B We know that,
de-Broglie wavelength, $λ = \frac{h}{\sqrt{2 mE}}$
$λ \propto \frac{1}{\sqrt{m}}$
$λ^{'} = λ \sqrt{\frac{m}{M}}$
Where, $M$ is the mass of the accelerated proton.

BITSAT-2006

Dual nature of radiation and Matter

142483 The minimum wavelength of $X$ - ray emitted from $X$ -ray machine operating at an accelerating potential of $V$ volts is:

1

\frac{hc}{eV}

2

\frac{Vc}{eh}

3

\frac{eh}{Vc}

4

\frac{eV}{hc}

Explanation:

A We know that,
$E_{max} = h v_{max} = eV$
$\frac{hc}{λ_{min}} = eV$
$λ_{min} = \frac{hc}{eV}$

UPSEE - 2005

Dual nature of radiation and Matter

142454 A particle of mass $m$ is projected from ground with velocity $u$ making angle $θ$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.

1

\infty

2

h / mu \sin θ

3

h / mu \cos θ

4

h / mu

Explanation:

B As we know that
de-Broglie wavelength $(λ) = \frac{h}{mv}$
At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin θ$
$∴$ the wavelength is given by, $λ = \frac{h}{mu \sin θ}$

AIIMS-26.05.2018(E)

Dual nature of radiation and Matter

142464 Which of the following figure represents the variation of particle momentum and associated de Broglie wavelength?

1

2

3

4

Explanation:

D According to the de-Broglie hypothesis wavelength, $λ = \frac{h}{p}$
$∴ λ \propto \frac{1}{p}$
From the above relegation we have rectangular hyperbola.

Hence, curve (d) is the correct option.

Karnataka CET-2006

Dual nature of radiation and Matter

142472 In Davission- Germer experiment maximum intensity is observed at

1

50^{\circ}

and

54 V

2

54^{\circ}

and

50 V

3

50^{\circ}

and

50 V

4

65^{\circ}

and

50 V

Explanation:

A The Davission - Germer experiment proves that an electron has dual nature i.e. it behaves as both wave and a particle. It shows that an electron beam could undergo diffraction when passed through the atomic crystal of nickel. In Davission - Germer experiments maximum intensity is observed at $50^{\circ}$ and $54^{\circ} V$ .
According to Bragg's law -
$n λ = 2 d \sin θ$
In Davission- Germer experiment maximum intensity is observed at $θ = 50^{\circ}$ and $V = 54 V$

MP PET-2008

Dual nature of radiation and Matter

142475 An electron of mass ' $m$ ', when accelerated through a potential $v$ has de-Broglie wavelength $λ$ . The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

λ (\frac{M}{m})

4

λ (\frac{m}{M})

Explanation:

B We know that,
de-Broglie wavelength, $λ = \frac{h}{\sqrt{2 mE}}$
$λ \propto \frac{1}{\sqrt{m}}$
$λ^{'} = λ \sqrt{\frac{m}{M}}$
Where, $M$ is the mass of the accelerated proton.

BITSAT-2006

Dual nature of radiation and Matter

142483 The minimum wavelength of $X$ - ray emitted from $X$ -ray machine operating at an accelerating potential of $V$ volts is:

1

\frac{hc}{eV}

2

\frac{Vc}{eh}

3

\frac{eh}{Vc}

4

\frac{eV}{hc}

Explanation:

A We know that,
$E_{max} = h v_{max} = eV$
$\frac{hc}{λ_{min}} = eV$
$λ_{min} = \frac{hc}{eV}$

UPSEE - 2005

Dual nature of radiation and Matter

142454 A particle of mass $m$ is projected from ground with velocity $u$ making angle $θ$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.

1

\infty

2

h / mu \sin θ

3

h / mu \cos θ

4

h / mu

Explanation:

B As we know that
de-Broglie wavelength $(λ) = \frac{h}{mv}$
At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin θ$
$∴$ the wavelength is given by, $λ = \frac{h}{mu \sin θ}$

AIIMS-26.05.2018(E)

Dual nature of radiation and Matter

142464 Which of the following figure represents the variation of particle momentum and associated de Broglie wavelength?

1

2

3

4

Explanation:

D According to the de-Broglie hypothesis wavelength, $λ = \frac{h}{p}$
$∴ λ \propto \frac{1}{p}$
From the above relegation we have rectangular hyperbola.

Hence, curve (d) is the correct option.

Karnataka CET-2006

Dual nature of radiation and Matter

142472 In Davission- Germer experiment maximum intensity is observed at

1

50^{\circ}

and

54 V

2

54^{\circ}

and

50 V

3

50^{\circ}

and

50 V

4

65^{\circ}

and

50 V

Explanation:

A The Davission - Germer experiment proves that an electron has dual nature i.e. it behaves as both wave and a particle. It shows that an electron beam could undergo diffraction when passed through the atomic crystal of nickel. In Davission - Germer experiments maximum intensity is observed at $50^{\circ}$ and $54^{\circ} V$ .
According to Bragg's law -
$n λ = 2 d \sin θ$
In Davission- Germer experiment maximum intensity is observed at $θ = 50^{\circ}$ and $V = 54 V$

MP PET-2008

Dual nature of radiation and Matter

142475 An electron of mass ' $m$ ', when accelerated through a potential $v$ has de-Broglie wavelength $λ$ . The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

λ (\frac{M}{m})

4

λ (\frac{m}{M})

Explanation:

B We know that,
de-Broglie wavelength, $λ = \frac{h}{\sqrt{2 mE}}$
$λ \propto \frac{1}{\sqrt{m}}$
$λ^{'} = λ \sqrt{\frac{m}{M}}$
Where, $M$ is the mass of the accelerated proton.

BITSAT-2006

Dual nature of radiation and Matter

142483 The minimum wavelength of $X$ - ray emitted from $X$ -ray machine operating at an accelerating potential of $V$ volts is:

1

\frac{hc}{eV}

2

\frac{Vc}{eh}

3

\frac{eh}{Vc}

4

\frac{eV}{hc}

Explanation:

A We know that,
$E_{max} = h v_{max} = eV$
$\frac{hc}{λ_{min}} = eV$
$λ_{min} = \frac{hc}{eV}$

UPSEE - 2005

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