Wave Nature of Matter
« Previous Topic: Photoelectric Effect Last Topic in »
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357905 Two particles are moving at right angle to each other, and their de Broglie wavelengths are \({\lambda_{1}}\) and \({\lambda_{2}}\), respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength \({\lambda}\) of the final particle is given by

1 \({\lambda=\dfrac{\lambda_{1}+\lambda_{2}}{2}}\)
2 \({\dfrac{2}{\lambda}=\dfrac{1}{\lambda_{1}}+\dfrac{1}{\lambda_{2}}}\)
3 \({\lambda=\sqrt{\lambda_{1} \lambda_{2}}}\)
4 \({\dfrac{1}{\lambda^{2}}=\dfrac{1}{\lambda_{1}^{2}}+\dfrac{1}{\lambda_{2}^{2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357906 A proton and an alpha particle are accelerated under the same potential difference. The ratio of de - Broglie wavelengths of the proton and the alpha particle is

1 \(\sqrt{8}\)
2 \(\dfrac{1}{\sqrt{8}}\)
3 1
4 2
KCET - 2012
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357907 A photon and electron have same de Broglie wavelength. Given that \(v\) is the speed of electron and \(c\) is the velocity of light. \(E_{e}, E_{p}\) are the kinetic energies of electron and photon respectively. \(p_{e}, p_{h}\) are the momentum of electron and photon respectively. Then which of the following relation is correct?

1 \(\frac{{{E_e}}}{{{E_p}}} = \frac{{2c}}{v}\)
2 \(\frac{{{P_e}}}{{{P_h}}} = \frac{{2c}}{v}\)
3 \(\frac{{{E_e}}}{{{E_p}}} = \frac{v}{{2c}}\)
4 \(\frac{{{P_e}}}{{{P_h}}} = \frac{c}{{2v}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357908 A proton, a neutron, an electron and an \(\alpha\)-particle have same energy. Then, their de Broglie wavelength compare as

1 \({\lambda _p} = {\lambda _n} > {\lambda _e} > {\lambda _\alpha }\)
2 \({\lambda _\alpha } < {\lambda _p} = {\lambda _n} < {\lambda _e}\)
3 \({\lambda _e} < {\lambda _p} = {\lambda _n} > {\lambda _\alpha }\)
4 \({\lambda _e} = {\lambda _p} = {\lambda _n} = {\lambda _\alpha }\)
NCERT Exempler
NEET DIGITAL LIBRARY
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357905 Two particles are moving at right angle to each other, and their de Broglie wavelengths are \({\lambda_{1}}\) and \({\lambda_{2}}\), respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength \({\lambda}\) of the final particle is given by

1 \({\lambda=\dfrac{\lambda_{1}+\lambda_{2}}{2}}\)
2 \({\dfrac{2}{\lambda}=\dfrac{1}{\lambda_{1}}+\dfrac{1}{\lambda_{2}}}\)
3 \({\lambda=\sqrt{\lambda_{1} \lambda_{2}}}\)
4 \({\dfrac{1}{\lambda^{2}}=\dfrac{1}{\lambda_{1}^{2}}+\dfrac{1}{\lambda_{2}^{2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357906 A proton and an alpha particle are accelerated under the same potential difference. The ratio of de - Broglie wavelengths of the proton and the alpha particle is

1 \(\sqrt{8}\)
2 \(\dfrac{1}{\sqrt{8}}\)
3 1
4 2
KCET - 2012
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357907 A photon and electron have same de Broglie wavelength. Given that \(v\) is the speed of electron and \(c\) is the velocity of light. \(E_{e}, E_{p}\) are the kinetic energies of electron and photon respectively. \(p_{e}, p_{h}\) are the momentum of electron and photon respectively. Then which of the following relation is correct?

1 \(\frac{{{E_e}}}{{{E_p}}} = \frac{{2c}}{v}\)
2 \(\frac{{{P_e}}}{{{P_h}}} = \frac{{2c}}{v}\)
3 \(\frac{{{E_e}}}{{{E_p}}} = \frac{v}{{2c}}\)
4 \(\frac{{{P_e}}}{{{P_h}}} = \frac{c}{{2v}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357908 A proton, a neutron, an electron and an \(\alpha\)-particle have same energy. Then, their de Broglie wavelength compare as

1 \({\lambda _p} = {\lambda _n} > {\lambda _e} > {\lambda _\alpha }\)
2 \({\lambda _\alpha } < {\lambda _p} = {\lambda _n} < {\lambda _e}\)
3 \({\lambda _e} < {\lambda _p} = {\lambda _n} > {\lambda _\alpha }\)
4 \({\lambda _e} = {\lambda _p} = {\lambda _n} = {\lambda _\alpha }\)
NCERT Exempler
Join With Us for More Updates
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357905 Two particles are moving at right angle to each other, and their de Broglie wavelengths are \({\lambda_{1}}\) and \({\lambda_{2}}\), respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength \({\lambda}\) of the final particle is given by

1 \({\lambda=\dfrac{\lambda_{1}+\lambda_{2}}{2}}\)
2 \({\dfrac{2}{\lambda}=\dfrac{1}{\lambda_{1}}+\dfrac{1}{\lambda_{2}}}\)
3 \({\lambda=\sqrt{\lambda_{1} \lambda_{2}}}\)
4 \({\dfrac{1}{\lambda^{2}}=\dfrac{1}{\lambda_{1}^{2}}+\dfrac{1}{\lambda_{2}^{2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357906 A proton and an alpha particle are accelerated under the same potential difference. The ratio of de - Broglie wavelengths of the proton and the alpha particle is

1 \(\sqrt{8}\)
2 \(\dfrac{1}{\sqrt{8}}\)
3 1
4 2
KCET - 2012
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357907 A photon and electron have same de Broglie wavelength. Given that \(v\) is the speed of electron and \(c\) is the velocity of light. \(E_{e}, E_{p}\) are the kinetic energies of electron and photon respectively. \(p_{e}, p_{h}\) are the momentum of electron and photon respectively. Then which of the following relation is correct?

1 \(\frac{{{E_e}}}{{{E_p}}} = \frac{{2c}}{v}\)
2 \(\frac{{{P_e}}}{{{P_h}}} = \frac{{2c}}{v}\)
3 \(\frac{{{E_e}}}{{{E_p}}} = \frac{v}{{2c}}\)
4 \(\frac{{{P_e}}}{{{P_h}}} = \frac{c}{{2v}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357908 A proton, a neutron, an electron and an \(\alpha\)-particle have same energy. Then, their de Broglie wavelength compare as

1 \({\lambda _p} = {\lambda _n} > {\lambda _e} > {\lambda _\alpha }\)
2 \({\lambda _\alpha } < {\lambda _p} = {\lambda _n} < {\lambda _e}\)
3 \({\lambda _e} < {\lambda _p} = {\lambda _n} > {\lambda _\alpha }\)
4 \({\lambda _e} = {\lambda _p} = {\lambda _n} = {\lambda _\alpha }\)
NCERT Exempler
For More Updates join with Us
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357905 Two particles are moving at right angle to each other, and their de Broglie wavelengths are \({\lambda_{1}}\) and \({\lambda_{2}}\), respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength \({\lambda}\) of the final particle is given by

1 \({\lambda=\dfrac{\lambda_{1}+\lambda_{2}}{2}}\)
2 \({\dfrac{2}{\lambda}=\dfrac{1}{\lambda_{1}}+\dfrac{1}{\lambda_{2}}}\)
3 \({\lambda=\sqrt{\lambda_{1} \lambda_{2}}}\)
4 \({\dfrac{1}{\lambda^{2}}=\dfrac{1}{\lambda_{1}^{2}}+\dfrac{1}{\lambda_{2}^{2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357906 A proton and an alpha particle are accelerated under the same potential difference. The ratio of de - Broglie wavelengths of the proton and the alpha particle is

1 \(\sqrt{8}\)
2 \(\dfrac{1}{\sqrt{8}}\)
3 1
4 2
KCET - 2012
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357907 A photon and electron have same de Broglie wavelength. Given that \(v\) is the speed of electron and \(c\) is the velocity of light. \(E_{e}, E_{p}\) are the kinetic energies of electron and photon respectively. \(p_{e}, p_{h}\) are the momentum of electron and photon respectively. Then which of the following relation is correct?

1 \(\frac{{{E_e}}}{{{E_p}}} = \frac{{2c}}{v}\)
2 \(\frac{{{P_e}}}{{{P_h}}} = \frac{{2c}}{v}\)
3 \(\frac{{{E_e}}}{{{E_p}}} = \frac{v}{{2c}}\)
4 \(\frac{{{P_e}}}{{{P_h}}} = \frac{c}{{2v}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357908 A proton, a neutron, an electron and an \(\alpha\)-particle have same energy. Then, their de Broglie wavelength compare as

1 \({\lambda _p} = {\lambda _n} > {\lambda _e} > {\lambda _\alpha }\)
2 \({\lambda _\alpha } < {\lambda _p} = {\lambda _n} < {\lambda _e}\)
3 \({\lambda _e} < {\lambda _p} = {\lambda _n} > {\lambda _\alpha }\)
4 \({\lambda _e} = {\lambda _p} = {\lambda _n} = {\lambda _\alpha }\)
NCERT Exempler
NEET DIGITAL LIBRARY