142523
The masses of two particles having same kinetic energies are in the ratio $2: 1$. Then their de Broglie wavelengths are in the ratio
1 $2: 1$
2 $1: 2$
3 $\sqrt{2}: 1$
4 $1: \sqrt{2}$
Explanation:
D We know that, de Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{K}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$ $\lambda_{1}: \lambda_{2}=1: \sqrt{2}$
J and K CET- 2009
Dual nature of radiation and Matter
142525
A particle with rest mass zero is moving with speed c. The de Broglie wavelength associated with it
1 zero
2 infinity
3 $\frac{h v}{c}$
4 $\frac{\mathrm{m}_{0} \mathrm{c}}{\mathrm{h}}$
Explanation:
A Theory of special relativity, $\mathrm{m}^{\prime}=\lambda \mathrm{m}=\frac{\mathrm{m}}{\sqrt{\frac{1-\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Where $\mathrm{m}^{\prime}=$ relative mass of particles $\mathrm{m}=$ mass of particles $\because \mathrm{v}=\mathrm{c}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{\sqrt{\frac{1-\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{0}=\infty$ It means the relative mass tend to Zero $\because$ de - Broglie wavelength is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Substitute the value, $\lambda=\frac{\mathrm{h}}{(\infty) \mathrm{c}}=0$ $\lambda=0$
J and K CET- 2004
Dual nature of radiation and Matter
142527
Consider the four gases- hydrogen, oxygen, nitrogen and helium at the same temperature Arrange them in the increasing order of the de Broglie wavelengths of their molecules
1 Hydrogen, helium, nitrogen, oxygen
2 Oxygen, nitrogen, hydrogen, helium
3 Oxygen, nitrogen, helium, hydrogen
4 Nitrogen, oxygen, helium, hydrogen
Explanation:
C We know that, $\lambda_{\text {gas }}=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mk}_{\mathrm{B}}^{\mathrm{T}}}}$ Where, $\mathrm{m}=$ mass of gas $\mathrm{k}_{\mathrm{B}}=\text { Boltzmann's constant }$ $\mathrm{T}=\text { Temperature }$ $\mathrm{O}, \mathrm{H}, \mathrm{He}, \mathrm{N}$ has constant temperature $\uparrow \lambda_{\text {gas }} \propto \frac{1}{\sqrt{\mathrm{m}} \downarrow}$ $\mathrm{m}_{\mathrm{O}}>\mathrm{m}_{\mathrm{N}}>\mathrm{m}_{\mathrm{He}}>\mathrm{m}_{\mathrm{H}}$ $\lambda_{\mathrm{O}} \lt \lambda_{\mathrm{N}} \lt \lambda_{\mathrm{He}} \lt \lambda_{\mathrm{H}}$
J and K CET-2012
Dual nature of radiation and Matter
142528
The de-Broglie wavelength of an electron is the same as that of a $50 \mathrm{keV} \mathrm{X}$-ray photon. The ratio of the energy if the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \mathrm{MeV}$ )
1 $1: 50$
2 $1: 20$
3 $20: 1$
4 $50: 1$
Explanation:
C Given $\mathrm{mc}^{2}=0.5 \mathrm{MeV}=0.5 \times 10^{6} \mathrm{eV}=5 \times 10^{5} \mathrm{eV}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=50 \mathrm{keV}=50 \times 10^{3} \mathrm{eV}=5 \times 10^{4} \mathrm{eV}$ We know that, de - Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ The kinetic energy of electron- $\mathrm{K}_{\text {electron }}=\frac{1}{2 \mathrm{~m}} \times \frac{\mathrm{h}^{2}}{\lambda^{2}}$ The photon energy - $\mathrm{E}_{\text {photon }}=\frac{\mathrm{hc}}{\lambda}$ Dividing equation (b) from equation (a) $\frac{\mathrm{E}_{\text {phothon }}}{\mathrm{K}_{\text {electron }}}=\frac{\frac{\mathrm{hc}}{\lambda}}{\frac{1}{2 \mathrm{~m}} \times \frac{\mathrm{h}^{2}}{\lambda^{2}}}=\frac{\mathrm{hc} \lambda^{2} \times 2 \mathrm{~m}}{\mathrm{~h}^{2} \times \lambda}$ $\frac{\mathrm{E}_{\text {phothon }}}{\mathrm{K}_{\text {electron }}}=\frac{2 \mathrm{~m} \lambda \mathrm{c}}{\mathrm{h}}=\frac{2 \mathrm{mc}^{2}}{\mathrm{hc} / \lambda}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=\frac{2 \times 5 \times 10^{5}}{5 \times 10^{4}}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=\frac{20}{1}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=20: 1$
142523
The masses of two particles having same kinetic energies are in the ratio $2: 1$. Then their de Broglie wavelengths are in the ratio
1 $2: 1$
2 $1: 2$
3 $\sqrt{2}: 1$
4 $1: \sqrt{2}$
Explanation:
D We know that, de Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{K}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$ $\lambda_{1}: \lambda_{2}=1: \sqrt{2}$
J and K CET- 2009
Dual nature of radiation and Matter
142525
A particle with rest mass zero is moving with speed c. The de Broglie wavelength associated with it
1 zero
2 infinity
3 $\frac{h v}{c}$
4 $\frac{\mathrm{m}_{0} \mathrm{c}}{\mathrm{h}}$
Explanation:
A Theory of special relativity, $\mathrm{m}^{\prime}=\lambda \mathrm{m}=\frac{\mathrm{m}}{\sqrt{\frac{1-\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Where $\mathrm{m}^{\prime}=$ relative mass of particles $\mathrm{m}=$ mass of particles $\because \mathrm{v}=\mathrm{c}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{\sqrt{\frac{1-\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{0}=\infty$ It means the relative mass tend to Zero $\because$ de - Broglie wavelength is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Substitute the value, $\lambda=\frac{\mathrm{h}}{(\infty) \mathrm{c}}=0$ $\lambda=0$
J and K CET- 2004
Dual nature of radiation and Matter
142527
Consider the four gases- hydrogen, oxygen, nitrogen and helium at the same temperature Arrange them in the increasing order of the de Broglie wavelengths of their molecules
1 Hydrogen, helium, nitrogen, oxygen
2 Oxygen, nitrogen, hydrogen, helium
3 Oxygen, nitrogen, helium, hydrogen
4 Nitrogen, oxygen, helium, hydrogen
Explanation:
C We know that, $\lambda_{\text {gas }}=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mk}_{\mathrm{B}}^{\mathrm{T}}}}$ Where, $\mathrm{m}=$ mass of gas $\mathrm{k}_{\mathrm{B}}=\text { Boltzmann's constant }$ $\mathrm{T}=\text { Temperature }$ $\mathrm{O}, \mathrm{H}, \mathrm{He}, \mathrm{N}$ has constant temperature $\uparrow \lambda_{\text {gas }} \propto \frac{1}{\sqrt{\mathrm{m}} \downarrow}$ $\mathrm{m}_{\mathrm{O}}>\mathrm{m}_{\mathrm{N}}>\mathrm{m}_{\mathrm{He}}>\mathrm{m}_{\mathrm{H}}$ $\lambda_{\mathrm{O}} \lt \lambda_{\mathrm{N}} \lt \lambda_{\mathrm{He}} \lt \lambda_{\mathrm{H}}$
J and K CET-2012
Dual nature of radiation and Matter
142528
The de-Broglie wavelength of an electron is the same as that of a $50 \mathrm{keV} \mathrm{X}$-ray photon. The ratio of the energy if the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \mathrm{MeV}$ )
1 $1: 50$
2 $1: 20$
3 $20: 1$
4 $50: 1$
Explanation:
C Given $\mathrm{mc}^{2}=0.5 \mathrm{MeV}=0.5 \times 10^{6} \mathrm{eV}=5 \times 10^{5} \mathrm{eV}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=50 \mathrm{keV}=50 \times 10^{3} \mathrm{eV}=5 \times 10^{4} \mathrm{eV}$ We know that, de - Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ The kinetic energy of electron- $\mathrm{K}_{\text {electron }}=\frac{1}{2 \mathrm{~m}} \times \frac{\mathrm{h}^{2}}{\lambda^{2}}$ The photon energy - $\mathrm{E}_{\text {photon }}=\frac{\mathrm{hc}}{\lambda}$ Dividing equation (b) from equation (a) $\frac{\mathrm{E}_{\text {phothon }}}{\mathrm{K}_{\text {electron }}}=\frac{\frac{\mathrm{hc}}{\lambda}}{\frac{1}{2 \mathrm{~m}} \times \frac{\mathrm{h}^{2}}{\lambda^{2}}}=\frac{\mathrm{hc} \lambda^{2} \times 2 \mathrm{~m}}{\mathrm{~h}^{2} \times \lambda}$ $\frac{\mathrm{E}_{\text {phothon }}}{\mathrm{K}_{\text {electron }}}=\frac{2 \mathrm{~m} \lambda \mathrm{c}}{\mathrm{h}}=\frac{2 \mathrm{mc}^{2}}{\mathrm{hc} / \lambda}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=\frac{2 \times 5 \times 10^{5}}{5 \times 10^{4}}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=\frac{20}{1}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=20: 1$
142523
The masses of two particles having same kinetic energies are in the ratio $2: 1$. Then their de Broglie wavelengths are in the ratio
1 $2: 1$
2 $1: 2$
3 $\sqrt{2}: 1$
4 $1: \sqrt{2}$
Explanation:
D We know that, de Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{K}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$ $\lambda_{1}: \lambda_{2}=1: \sqrt{2}$
J and K CET- 2009
Dual nature of radiation and Matter
142525
A particle with rest mass zero is moving with speed c. The de Broglie wavelength associated with it
1 zero
2 infinity
3 $\frac{h v}{c}$
4 $\frac{\mathrm{m}_{0} \mathrm{c}}{\mathrm{h}}$
Explanation:
A Theory of special relativity, $\mathrm{m}^{\prime}=\lambda \mathrm{m}=\frac{\mathrm{m}}{\sqrt{\frac{1-\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Where $\mathrm{m}^{\prime}=$ relative mass of particles $\mathrm{m}=$ mass of particles $\because \mathrm{v}=\mathrm{c}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{\sqrt{\frac{1-\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{0}=\infty$ It means the relative mass tend to Zero $\because$ de - Broglie wavelength is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Substitute the value, $\lambda=\frac{\mathrm{h}}{(\infty) \mathrm{c}}=0$ $\lambda=0$
J and K CET- 2004
Dual nature of radiation and Matter
142527
Consider the four gases- hydrogen, oxygen, nitrogen and helium at the same temperature Arrange them in the increasing order of the de Broglie wavelengths of their molecules
1 Hydrogen, helium, nitrogen, oxygen
2 Oxygen, nitrogen, hydrogen, helium
3 Oxygen, nitrogen, helium, hydrogen
4 Nitrogen, oxygen, helium, hydrogen
Explanation:
C We know that, $\lambda_{\text {gas }}=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mk}_{\mathrm{B}}^{\mathrm{T}}}}$ Where, $\mathrm{m}=$ mass of gas $\mathrm{k}_{\mathrm{B}}=\text { Boltzmann's constant }$ $\mathrm{T}=\text { Temperature }$ $\mathrm{O}, \mathrm{H}, \mathrm{He}, \mathrm{N}$ has constant temperature $\uparrow \lambda_{\text {gas }} \propto \frac{1}{\sqrt{\mathrm{m}} \downarrow}$ $\mathrm{m}_{\mathrm{O}}>\mathrm{m}_{\mathrm{N}}>\mathrm{m}_{\mathrm{He}}>\mathrm{m}_{\mathrm{H}}$ $\lambda_{\mathrm{O}} \lt \lambda_{\mathrm{N}} \lt \lambda_{\mathrm{He}} \lt \lambda_{\mathrm{H}}$
J and K CET-2012
Dual nature of radiation and Matter
142528
The de-Broglie wavelength of an electron is the same as that of a $50 \mathrm{keV} \mathrm{X}$-ray photon. The ratio of the energy if the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \mathrm{MeV}$ )
1 $1: 50$
2 $1: 20$
3 $20: 1$
4 $50: 1$
Explanation:
C Given $\mathrm{mc}^{2}=0.5 \mathrm{MeV}=0.5 \times 10^{6} \mathrm{eV}=5 \times 10^{5} \mathrm{eV}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=50 \mathrm{keV}=50 \times 10^{3} \mathrm{eV}=5 \times 10^{4} \mathrm{eV}$ We know that, de - Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ The kinetic energy of electron- $\mathrm{K}_{\text {electron }}=\frac{1}{2 \mathrm{~m}} \times \frac{\mathrm{h}^{2}}{\lambda^{2}}$ The photon energy - $\mathrm{E}_{\text {photon }}=\frac{\mathrm{hc}}{\lambda}$ Dividing equation (b) from equation (a) $\frac{\mathrm{E}_{\text {phothon }}}{\mathrm{K}_{\text {electron }}}=\frac{\frac{\mathrm{hc}}{\lambda}}{\frac{1}{2 \mathrm{~m}} \times \frac{\mathrm{h}^{2}}{\lambda^{2}}}=\frac{\mathrm{hc} \lambda^{2} \times 2 \mathrm{~m}}{\mathrm{~h}^{2} \times \lambda}$ $\frac{\mathrm{E}_{\text {phothon }}}{\mathrm{K}_{\text {electron }}}=\frac{2 \mathrm{~m} \lambda \mathrm{c}}{\mathrm{h}}=\frac{2 \mathrm{mc}^{2}}{\mathrm{hc} / \lambda}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=\frac{2 \times 5 \times 10^{5}}{5 \times 10^{4}}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=\frac{20}{1}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=20: 1$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142523
The masses of two particles having same kinetic energies are in the ratio $2: 1$. Then their de Broglie wavelengths are in the ratio
1 $2: 1$
2 $1: 2$
3 $\sqrt{2}: 1$
4 $1: \sqrt{2}$
Explanation:
D We know that, de Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{K}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$ $\lambda_{1}: \lambda_{2}=1: \sqrt{2}$
J and K CET- 2009
Dual nature of radiation and Matter
142525
A particle with rest mass zero is moving with speed c. The de Broglie wavelength associated with it
1 zero
2 infinity
3 $\frac{h v}{c}$
4 $\frac{\mathrm{m}_{0} \mathrm{c}}{\mathrm{h}}$
Explanation:
A Theory of special relativity, $\mathrm{m}^{\prime}=\lambda \mathrm{m}=\frac{\mathrm{m}}{\sqrt{\frac{1-\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Where $\mathrm{m}^{\prime}=$ relative mass of particles $\mathrm{m}=$ mass of particles $\because \mathrm{v}=\mathrm{c}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{\sqrt{\frac{1-\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{0}=\infty$ It means the relative mass tend to Zero $\because$ de - Broglie wavelength is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Substitute the value, $\lambda=\frac{\mathrm{h}}{(\infty) \mathrm{c}}=0$ $\lambda=0$
J and K CET- 2004
Dual nature of radiation and Matter
142527
Consider the four gases- hydrogen, oxygen, nitrogen and helium at the same temperature Arrange them in the increasing order of the de Broglie wavelengths of their molecules
1 Hydrogen, helium, nitrogen, oxygen
2 Oxygen, nitrogen, hydrogen, helium
3 Oxygen, nitrogen, helium, hydrogen
4 Nitrogen, oxygen, helium, hydrogen
Explanation:
C We know that, $\lambda_{\text {gas }}=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mk}_{\mathrm{B}}^{\mathrm{T}}}}$ Where, $\mathrm{m}=$ mass of gas $\mathrm{k}_{\mathrm{B}}=\text { Boltzmann's constant }$ $\mathrm{T}=\text { Temperature }$ $\mathrm{O}, \mathrm{H}, \mathrm{He}, \mathrm{N}$ has constant temperature $\uparrow \lambda_{\text {gas }} \propto \frac{1}{\sqrt{\mathrm{m}} \downarrow}$ $\mathrm{m}_{\mathrm{O}}>\mathrm{m}_{\mathrm{N}}>\mathrm{m}_{\mathrm{He}}>\mathrm{m}_{\mathrm{H}}$ $\lambda_{\mathrm{O}} \lt \lambda_{\mathrm{N}} \lt \lambda_{\mathrm{He}} \lt \lambda_{\mathrm{H}}$
J and K CET-2012
Dual nature of radiation and Matter
142528
The de-Broglie wavelength of an electron is the same as that of a $50 \mathrm{keV} \mathrm{X}$-ray photon. The ratio of the energy if the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \mathrm{MeV}$ )
1 $1: 50$
2 $1: 20$
3 $20: 1$
4 $50: 1$
Explanation:
C Given $\mathrm{mc}^{2}=0.5 \mathrm{MeV}=0.5 \times 10^{6} \mathrm{eV}=5 \times 10^{5} \mathrm{eV}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=50 \mathrm{keV}=50 \times 10^{3} \mathrm{eV}=5 \times 10^{4} \mathrm{eV}$ We know that, de - Broglie wavelength $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ The kinetic energy of electron- $\mathrm{K}_{\text {electron }}=\frac{1}{2 \mathrm{~m}} \times \frac{\mathrm{h}^{2}}{\lambda^{2}}$ The photon energy - $\mathrm{E}_{\text {photon }}=\frac{\mathrm{hc}}{\lambda}$ Dividing equation (b) from equation (a) $\frac{\mathrm{E}_{\text {phothon }}}{\mathrm{K}_{\text {electron }}}=\frac{\frac{\mathrm{hc}}{\lambda}}{\frac{1}{2 \mathrm{~m}} \times \frac{\mathrm{h}^{2}}{\lambda^{2}}}=\frac{\mathrm{hc} \lambda^{2} \times 2 \mathrm{~m}}{\mathrm{~h}^{2} \times \lambda}$ $\frac{\mathrm{E}_{\text {phothon }}}{\mathrm{K}_{\text {electron }}}=\frac{2 \mathrm{~m} \lambda \mathrm{c}}{\mathrm{h}}=\frac{2 \mathrm{mc}^{2}}{\mathrm{hc} / \lambda}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=\frac{2 \times 5 \times 10^{5}}{5 \times 10^{4}}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=\frac{20}{1}$ $\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{K}_{\mathrm{e}}}=20: 1$
