142454
A particle of mass $m$ is projected from ground with velocity $u$ making angle $\theta$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.
1 $\infty$
2 $\mathrm{h} / \mathrm{mu} \sin \theta$
3 $\mathrm{h} / \mathrm{mu} \cos \theta$
4 $\mathrm{h} / \mathrm{mu}$
Explanation:
B As we know that de-Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\mathrm{mv}}$ At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin \theta$ $\therefore$ the wavelength is given by, $\lambda=\frac{\mathrm{h}}{\mathrm{mu} \sin \theta}$
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, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\therefore \quad \lambda \propto \frac{1}{\mathrm{p}}$ 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 $54 \mathrm{~V}$
2 $54^{\circ}$ and $50 \mathrm{~V}$
3 $50^{\circ}$ and $50 \mathrm{~V}$
4 $65^{\circ}$ and $50 \mathrm{~V}$
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 $54^{\circ} \mathrm{V}$. According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ In Davission- Germer experiment maximum intensity is observed at $\theta=50^{\circ}$ and $\mathrm{V}=54 \mathrm{~V}$
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 $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -
B We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda^{\prime}=\lambda \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$ Where, $\mathrm{M}$ 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 $\frac{\mathrm{hc}}{\mathrm{eV}}$
2 $\frac{\mathrm{Vc}}{\mathrm{eh}}$
3 $\frac{\mathrm{eh}}{\mathrm{Vc}}$
4 $\frac{\mathrm{eV}}{\mathrm{hc}}$
Explanation:
A We know that, $\mathrm{E}_{\max }=\mathrm{h} v_{\max }=\mathrm{eV}$ $\frac{\mathrm{hc}}{\lambda_{\text {min }}}=\mathrm{eV}$ $\lambda_{\text {min }}=\frac{\mathrm{hc}}{\mathrm{eV}}$
142454
A particle of mass $m$ is projected from ground with velocity $u$ making angle $\theta$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.
1 $\infty$
2 $\mathrm{h} / \mathrm{mu} \sin \theta$
3 $\mathrm{h} / \mathrm{mu} \cos \theta$
4 $\mathrm{h} / \mathrm{mu}$
Explanation:
B As we know that de-Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\mathrm{mv}}$ At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin \theta$ $\therefore$ the wavelength is given by, $\lambda=\frac{\mathrm{h}}{\mathrm{mu} \sin \theta}$
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, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\therefore \quad \lambda \propto \frac{1}{\mathrm{p}}$ 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 $54 \mathrm{~V}$
2 $54^{\circ}$ and $50 \mathrm{~V}$
3 $50^{\circ}$ and $50 \mathrm{~V}$
4 $65^{\circ}$ and $50 \mathrm{~V}$
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 $54^{\circ} \mathrm{V}$. According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ In Davission- Germer experiment maximum intensity is observed at $\theta=50^{\circ}$ and $\mathrm{V}=54 \mathrm{~V}$
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 $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -
B We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda^{\prime}=\lambda \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$ Where, $\mathrm{M}$ 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 $\frac{\mathrm{hc}}{\mathrm{eV}}$
2 $\frac{\mathrm{Vc}}{\mathrm{eh}}$
3 $\frac{\mathrm{eh}}{\mathrm{Vc}}$
4 $\frac{\mathrm{eV}}{\mathrm{hc}}$
Explanation:
A We know that, $\mathrm{E}_{\max }=\mathrm{h} v_{\max }=\mathrm{eV}$ $\frac{\mathrm{hc}}{\lambda_{\text {min }}}=\mathrm{eV}$ $\lambda_{\text {min }}=\frac{\mathrm{hc}}{\mathrm{eV}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142454
A particle of mass $m$ is projected from ground with velocity $u$ making angle $\theta$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.
1 $\infty$
2 $\mathrm{h} / \mathrm{mu} \sin \theta$
3 $\mathrm{h} / \mathrm{mu} \cos \theta$
4 $\mathrm{h} / \mathrm{mu}$
Explanation:
B As we know that de-Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\mathrm{mv}}$ At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin \theta$ $\therefore$ the wavelength is given by, $\lambda=\frac{\mathrm{h}}{\mathrm{mu} \sin \theta}$
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, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\therefore \quad \lambda \propto \frac{1}{\mathrm{p}}$ 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 $54 \mathrm{~V}$
2 $54^{\circ}$ and $50 \mathrm{~V}$
3 $50^{\circ}$ and $50 \mathrm{~V}$
4 $65^{\circ}$ and $50 \mathrm{~V}$
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 $54^{\circ} \mathrm{V}$. According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ In Davission- Germer experiment maximum intensity is observed at $\theta=50^{\circ}$ and $\mathrm{V}=54 \mathrm{~V}$
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 $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -
B We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda^{\prime}=\lambda \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$ Where, $\mathrm{M}$ 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 $\frac{\mathrm{hc}}{\mathrm{eV}}$
2 $\frac{\mathrm{Vc}}{\mathrm{eh}}$
3 $\frac{\mathrm{eh}}{\mathrm{Vc}}$
4 $\frac{\mathrm{eV}}{\mathrm{hc}}$
Explanation:
A We know that, $\mathrm{E}_{\max }=\mathrm{h} v_{\max }=\mathrm{eV}$ $\frac{\mathrm{hc}}{\lambda_{\text {min }}}=\mathrm{eV}$ $\lambda_{\text {min }}=\frac{\mathrm{hc}}{\mathrm{eV}}$
142454
A particle of mass $m$ is projected from ground with velocity $u$ making angle $\theta$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.
1 $\infty$
2 $\mathrm{h} / \mathrm{mu} \sin \theta$
3 $\mathrm{h} / \mathrm{mu} \cos \theta$
4 $\mathrm{h} / \mathrm{mu}$
Explanation:
B As we know that de-Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\mathrm{mv}}$ At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin \theta$ $\therefore$ the wavelength is given by, $\lambda=\frac{\mathrm{h}}{\mathrm{mu} \sin \theta}$
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, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\therefore \quad \lambda \propto \frac{1}{\mathrm{p}}$ 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 $54 \mathrm{~V}$
2 $54^{\circ}$ and $50 \mathrm{~V}$
3 $50^{\circ}$ and $50 \mathrm{~V}$
4 $65^{\circ}$ and $50 \mathrm{~V}$
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 $54^{\circ} \mathrm{V}$. According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ In Davission- Germer experiment maximum intensity is observed at $\theta=50^{\circ}$ and $\mathrm{V}=54 \mathrm{~V}$
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 $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -
B We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda^{\prime}=\lambda \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$ Where, $\mathrm{M}$ 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 $\frac{\mathrm{hc}}{\mathrm{eV}}$
2 $\frac{\mathrm{Vc}}{\mathrm{eh}}$
3 $\frac{\mathrm{eh}}{\mathrm{Vc}}$
4 $\frac{\mathrm{eV}}{\mathrm{hc}}$
Explanation:
A We know that, $\mathrm{E}_{\max }=\mathrm{h} v_{\max }=\mathrm{eV}$ $\frac{\mathrm{hc}}{\lambda_{\text {min }}}=\mathrm{eV}$ $\lambda_{\text {min }}=\frac{\mathrm{hc}}{\mathrm{eV}}$
142454
A particle of mass $m$ is projected from ground with velocity $u$ making angle $\theta$ with the vertical. The de-Broglie wavelength of the particle at the highest point is.
1 $\infty$
2 $\mathrm{h} / \mathrm{mu} \sin \theta$
3 $\mathrm{h} / \mathrm{mu} \cos \theta$
4 $\mathrm{h} / \mathrm{mu}$
Explanation:
B As we know that de-Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\mathrm{mv}}$ At the highest point, only the horizontal component of the velocity will be there, i.e u $\sin \theta$ $\therefore$ the wavelength is given by, $\lambda=\frac{\mathrm{h}}{\mathrm{mu} \sin \theta}$
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, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\therefore \quad \lambda \propto \frac{1}{\mathrm{p}}$ 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 $54 \mathrm{~V}$
2 $54^{\circ}$ and $50 \mathrm{~V}$
3 $50^{\circ}$ and $50 \mathrm{~V}$
4 $65^{\circ}$ and $50 \mathrm{~V}$
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 $54^{\circ} \mathrm{V}$. According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ In Davission- Germer experiment maximum intensity is observed at $\theta=50^{\circ}$ and $\mathrm{V}=54 \mathrm{~V}$
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 $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be -
B We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda^{\prime}=\lambda \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$ Where, $\mathrm{M}$ 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 $\frac{\mathrm{hc}}{\mathrm{eV}}$
2 $\frac{\mathrm{Vc}}{\mathrm{eh}}$
3 $\frac{\mathrm{eh}}{\mathrm{Vc}}$
4 $\frac{\mathrm{eV}}{\mathrm{hc}}$
Explanation:
A We know that, $\mathrm{E}_{\max }=\mathrm{h} v_{\max }=\mathrm{eV}$ $\frac{\mathrm{hc}}{\lambda_{\text {min }}}=\mathrm{eV}$ $\lambda_{\text {min }}=\frac{\mathrm{hc}}{\mathrm{eV}}$
