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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357944 The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of the photon and the momentum of electron is (\(c\) = velocity of light, \(h\) = Planck's constant):

1 \(h\)
2 \(c\)
3 \(\frac{1}{h}\)
4 \(\frac{1}{c}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357945 An electron (mass \(m\) ) with an initial velocity \(v = {v_0}\hat i\left( {{v_0} > 0} \right)\) in an electric field \(E=E_{0} \hat{i}\) \(\left(E_{0}=\right.\) constant \(\left.>0\right)\). It's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0}\)
2 \(\lambda_{0} t\)
3 \(\frac{{{\lambda _0}}}{{m\left( {1 + \frac{{e{E_0}t}}{{m{v_0}}}} \right)}}\)
4 \({\lambda _0}\left( {1 + \frac{{e{E_0}t}}{{m{v_0}}}} \right)\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357946 Moving with the same velocity, one of the following has the longest de Broglie wavelength

1 \(\beta\)-particle
2 \(\alpha\)-particle
3 Proton
4 Neutron
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357947 If we express the energy of a photon in \(KeV\) and the wavelength in angstroms, then energy of a photon can be calculated from the relation

1 \(E=12.4 h / \lambda\)
2 \(E = 12.4\,hv\)
3 \(E=h v\)
4 \(E=12.4 / \lambda\)
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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357944 The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of the photon and the momentum of electron is (\(c\) = velocity of light, \(h\) = Planck's constant):

1 \(h\)
2 \(c\)
3 \(\frac{1}{h}\)
4 \(\frac{1}{c}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357945 An electron (mass \(m\) ) with an initial velocity \(v = {v_0}\hat i\left( {{v_0} > 0} \right)\) in an electric field \(E=E_{0} \hat{i}\) \(\left(E_{0}=\right.\) constant \(\left.>0\right)\). It's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0}\)
2 \(\lambda_{0} t\)
3 \(\frac{{{\lambda _0}}}{{m\left( {1 + \frac{{e{E_0}t}}{{m{v_0}}}} \right)}}\)
4 \({\lambda _0}\left( {1 + \frac{{e{E_0}t}}{{m{v_0}}}} \right)\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357946 Moving with the same velocity, one of the following has the longest de Broglie wavelength

1 \(\beta\)-particle
2 \(\alpha\)-particle
3 Proton
4 Neutron
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357947 If we express the energy of a photon in \(KeV\) and the wavelength in angstroms, then energy of a photon can be calculated from the relation

1 \(E=12.4 h / \lambda\)
2 \(E = 12.4\,hv\)
3 \(E=h v\)
4 \(E=12.4 / \lambda\)
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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357944 The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of the photon and the momentum of electron is (\(c\) = velocity of light, \(h\) = Planck's constant):

1 \(h\)
2 \(c\)
3 \(\frac{1}{h}\)
4 \(\frac{1}{c}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357945 An electron (mass \(m\) ) with an initial velocity \(v = {v_0}\hat i\left( {{v_0} > 0} \right)\) in an electric field \(E=E_{0} \hat{i}\) \(\left(E_{0}=\right.\) constant \(\left.>0\right)\). It's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0}\)
2 \(\lambda_{0} t\)
3 \(\frac{{{\lambda _0}}}{{m\left( {1 + \frac{{e{E_0}t}}{{m{v_0}}}} \right)}}\)
4 \({\lambda _0}\left( {1 + \frac{{e{E_0}t}}{{m{v_0}}}} \right)\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357946 Moving with the same velocity, one of the following has the longest de Broglie wavelength

1 \(\beta\)-particle
2 \(\alpha\)-particle
3 Proton
4 Neutron
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357947 If we express the energy of a photon in \(KeV\) and the wavelength in angstroms, then energy of a photon can be calculated from the relation

1 \(E=12.4 h / \lambda\)
2 \(E = 12.4\,hv\)
3 \(E=h v\)
4 \(E=12.4 / \lambda\)
For More Updates join with Us
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357944 The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of the photon and the momentum of electron is (\(c\) = velocity of light, \(h\) = Planck's constant):

1 \(h\)
2 \(c\)
3 \(\frac{1}{h}\)
4 \(\frac{1}{c}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357945 An electron (mass \(m\) ) with an initial velocity \(v = {v_0}\hat i\left( {{v_0} > 0} \right)\) in an electric field \(E=E_{0} \hat{i}\) \(\left(E_{0}=\right.\) constant \(\left.>0\right)\). It's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0}\)
2 \(\lambda_{0} t\)
3 \(\frac{{{\lambda _0}}}{{m\left( {1 + \frac{{e{E_0}t}}{{m{v_0}}}} \right)}}\)
4 \({\lambda _0}\left( {1 + \frac{{e{E_0}t}}{{m{v_0}}}} \right)\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357946 Moving with the same velocity, one of the following has the longest de Broglie wavelength

1 \(\beta\)-particle
2 \(\alpha\)-particle
3 Proton
4 Neutron
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357947 If we express the energy of a photon in \(KeV\) and the wavelength in angstroms, then energy of a photon can be calculated from the relation

1 \(E=12.4 h / \lambda\)
2 \(E = 12.4\,hv\)
3 \(E=h v\)
4 \(E=12.4 / \lambda\)