357935
Assertion : Both interference and diffraction shown by electrons are due to concept of matter wave. Reason : Electrons moving at proper velocities have their matter waves associated with themselves.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Matter waves, introduced by de Broglie, describe the dual nature of particles, explaining quantum phenomenae. So correct option is (1).
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357936
The curve drawn between velocity and frequency of photon in vacuum will be a
1 Straight line parallel to frequency axis
2 Straight line parallel to velocity axis
3 Straight line passing through origin and making an angle of \(45^{\circ}\) with frequency axis
4 Hyperbola
Explanation:
The curve drawn between velocity \((v)\) and frequency \((f)\) of photon in vacuum will be a straight line parallel to frequency axis as velocity of photon in a vacuum is constant and it is independent of frequency.
AIIMS - 2009
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357937
The de Broglie wavelength associated with the electron in the \(n = 4\) level is :
1 \({\frac{1}{4}^{th}}\) of the de Broglie wavelength of the electron in the ground state
2 Four times the de Broglie wavelength of the electron in the ground state
3 Two times the de Broglie wavelength of the electron in the ground state
4 Half of the de Broglie wavelength of the electron in the ground state
Explanation:
de Broglie wavelength of electron \(\lambda = \frac{h}{{mv}}\) As we know, \(v\,\alpha \frac{1}{n}\) So, \(\lambda \,\alpha \,n \Rightarrow {\lambda _4} = 4{\lambda _1}\) \(\lambda_{1}\) is the de Broglie wavelength of the electron in the ground state.
JEE - 2015
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357938
An electromagnetic wave of wavelength ' \(\lambda\) ', is incident on a photosensitive surface of negligible work function. If ' \(m\) ' is the mass of photoelectron emitted from the surface has deBroglie wavelength ' \(\lambda_{d}\) ' then
\(\dfrac{h c}{\lambda}=\dfrac{1}{2} m v^{2}+\phi\) Given that \(\phi \approx 0\) \(\dfrac{h c}{\lambda}=\dfrac{1}{2} m v^{2}\) \(\begin{aligned}& \dfrac{h c}{\lambda}=\dfrac{P^{2}}{2 m} \\& P=\sqrt{\dfrac{2 m h c}{\lambda}} \\& \lambda_{d}=\dfrac{h}{p}=\dfrac{h}{\sqrt{\dfrac{2 m h c}{\lambda}}} \\& =\sqrt{\dfrac{h^{2} \lambda}{2 m h c}}=\sqrt{\dfrac{h \lambda}{2 m c}} \\& \lambda_{d}^{2}=\dfrac{h \lambda}{2 m c} \Rightarrow \lambda=\dfrac{2 m c}{h} \lambda_{d}^{2}\end{aligned}\)
357935
Assertion : Both interference and diffraction shown by electrons are due to concept of matter wave. Reason : Electrons moving at proper velocities have their matter waves associated with themselves.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Matter waves, introduced by de Broglie, describe the dual nature of particles, explaining quantum phenomenae. So correct option is (1).
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357936
The curve drawn between velocity and frequency of photon in vacuum will be a
1 Straight line parallel to frequency axis
2 Straight line parallel to velocity axis
3 Straight line passing through origin and making an angle of \(45^{\circ}\) with frequency axis
4 Hyperbola
Explanation:
The curve drawn between velocity \((v)\) and frequency \((f)\) of photon in vacuum will be a straight line parallel to frequency axis as velocity of photon in a vacuum is constant and it is independent of frequency.
AIIMS - 2009
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357937
The de Broglie wavelength associated with the electron in the \(n = 4\) level is :
1 \({\frac{1}{4}^{th}}\) of the de Broglie wavelength of the electron in the ground state
2 Four times the de Broglie wavelength of the electron in the ground state
3 Two times the de Broglie wavelength of the electron in the ground state
4 Half of the de Broglie wavelength of the electron in the ground state
Explanation:
de Broglie wavelength of electron \(\lambda = \frac{h}{{mv}}\) As we know, \(v\,\alpha \frac{1}{n}\) So, \(\lambda \,\alpha \,n \Rightarrow {\lambda _4} = 4{\lambda _1}\) \(\lambda_{1}\) is the de Broglie wavelength of the electron in the ground state.
JEE - 2015
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357938
An electromagnetic wave of wavelength ' \(\lambda\) ', is incident on a photosensitive surface of negligible work function. If ' \(m\) ' is the mass of photoelectron emitted from the surface has deBroglie wavelength ' \(\lambda_{d}\) ' then
\(\dfrac{h c}{\lambda}=\dfrac{1}{2} m v^{2}+\phi\) Given that \(\phi \approx 0\) \(\dfrac{h c}{\lambda}=\dfrac{1}{2} m v^{2}\) \(\begin{aligned}& \dfrac{h c}{\lambda}=\dfrac{P^{2}}{2 m} \\& P=\sqrt{\dfrac{2 m h c}{\lambda}} \\& \lambda_{d}=\dfrac{h}{p}=\dfrac{h}{\sqrt{\dfrac{2 m h c}{\lambda}}} \\& =\sqrt{\dfrac{h^{2} \lambda}{2 m h c}}=\sqrt{\dfrac{h \lambda}{2 m c}} \\& \lambda_{d}^{2}=\dfrac{h \lambda}{2 m c} \Rightarrow \lambda=\dfrac{2 m c}{h} \lambda_{d}^{2}\end{aligned}\)
357935
Assertion : Both interference and diffraction shown by electrons are due to concept of matter wave. Reason : Electrons moving at proper velocities have their matter waves associated with themselves.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Matter waves, introduced by de Broglie, describe the dual nature of particles, explaining quantum phenomenae. So correct option is (1).
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357936
The curve drawn between velocity and frequency of photon in vacuum will be a
1 Straight line parallel to frequency axis
2 Straight line parallel to velocity axis
3 Straight line passing through origin and making an angle of \(45^{\circ}\) with frequency axis
4 Hyperbola
Explanation:
The curve drawn between velocity \((v)\) and frequency \((f)\) of photon in vacuum will be a straight line parallel to frequency axis as velocity of photon in a vacuum is constant and it is independent of frequency.
AIIMS - 2009
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357937
The de Broglie wavelength associated with the electron in the \(n = 4\) level is :
1 \({\frac{1}{4}^{th}}\) of the de Broglie wavelength of the electron in the ground state
2 Four times the de Broglie wavelength of the electron in the ground state
3 Two times the de Broglie wavelength of the electron in the ground state
4 Half of the de Broglie wavelength of the electron in the ground state
Explanation:
de Broglie wavelength of electron \(\lambda = \frac{h}{{mv}}\) As we know, \(v\,\alpha \frac{1}{n}\) So, \(\lambda \,\alpha \,n \Rightarrow {\lambda _4} = 4{\lambda _1}\) \(\lambda_{1}\) is the de Broglie wavelength of the electron in the ground state.
JEE - 2015
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357938
An electromagnetic wave of wavelength ' \(\lambda\) ', is incident on a photosensitive surface of negligible work function. If ' \(m\) ' is the mass of photoelectron emitted from the surface has deBroglie wavelength ' \(\lambda_{d}\) ' then
\(\dfrac{h c}{\lambda}=\dfrac{1}{2} m v^{2}+\phi\) Given that \(\phi \approx 0\) \(\dfrac{h c}{\lambda}=\dfrac{1}{2} m v^{2}\) \(\begin{aligned}& \dfrac{h c}{\lambda}=\dfrac{P^{2}}{2 m} \\& P=\sqrt{\dfrac{2 m h c}{\lambda}} \\& \lambda_{d}=\dfrac{h}{p}=\dfrac{h}{\sqrt{\dfrac{2 m h c}{\lambda}}} \\& =\sqrt{\dfrac{h^{2} \lambda}{2 m h c}}=\sqrt{\dfrac{h \lambda}{2 m c}} \\& \lambda_{d}^{2}=\dfrac{h \lambda}{2 m c} \Rightarrow \lambda=\dfrac{2 m c}{h} \lambda_{d}^{2}\end{aligned}\)
357935
Assertion : Both interference and diffraction shown by electrons are due to concept of matter wave. Reason : Electrons moving at proper velocities have their matter waves associated with themselves.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Matter waves, introduced by de Broglie, describe the dual nature of particles, explaining quantum phenomenae. So correct option is (1).
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357936
The curve drawn between velocity and frequency of photon in vacuum will be a
1 Straight line parallel to frequency axis
2 Straight line parallel to velocity axis
3 Straight line passing through origin and making an angle of \(45^{\circ}\) with frequency axis
4 Hyperbola
Explanation:
The curve drawn between velocity \((v)\) and frequency \((f)\) of photon in vacuum will be a straight line parallel to frequency axis as velocity of photon in a vacuum is constant and it is independent of frequency.
AIIMS - 2009
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357937
The de Broglie wavelength associated with the electron in the \(n = 4\) level is :
1 \({\frac{1}{4}^{th}}\) of the de Broglie wavelength of the electron in the ground state
2 Four times the de Broglie wavelength of the electron in the ground state
3 Two times the de Broglie wavelength of the electron in the ground state
4 Half of the de Broglie wavelength of the electron in the ground state
Explanation:
de Broglie wavelength of electron \(\lambda = \frac{h}{{mv}}\) As we know, \(v\,\alpha \frac{1}{n}\) So, \(\lambda \,\alpha \,n \Rightarrow {\lambda _4} = 4{\lambda _1}\) \(\lambda_{1}\) is the de Broglie wavelength of the electron in the ground state.
JEE - 2015
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357938
An electromagnetic wave of wavelength ' \(\lambda\) ', is incident on a photosensitive surface of negligible work function. If ' \(m\) ' is the mass of photoelectron emitted from the surface has deBroglie wavelength ' \(\lambda_{d}\) ' then
\(\dfrac{h c}{\lambda}=\dfrac{1}{2} m v^{2}+\phi\) Given that \(\phi \approx 0\) \(\dfrac{h c}{\lambda}=\dfrac{1}{2} m v^{2}\) \(\begin{aligned}& \dfrac{h c}{\lambda}=\dfrac{P^{2}}{2 m} \\& P=\sqrt{\dfrac{2 m h c}{\lambda}} \\& \lambda_{d}=\dfrac{h}{p}=\dfrac{h}{\sqrt{\dfrac{2 m h c}{\lambda}}} \\& =\sqrt{\dfrac{h^{2} \lambda}{2 m h c}}=\sqrt{\dfrac{h \lambda}{2 m c}} \\& \lambda_{d}^{2}=\dfrac{h \lambda}{2 m c} \Rightarrow \lambda=\dfrac{2 m c}{h} \lambda_{d}^{2}\end{aligned}\)
