357931
Which of the following particles - neutron, proton, electron and deuteron has the lowest energy if all have the same de Broglie wavelength?
1 Neutron
2 Proton
3 Electron
4 Deuteron
Explanation:
\(\lambda=\dfrac{h}{\sqrt{2 m E}} \Rightarrow E=\dfrac{h^{2}}{2 m \lambda^{2}}\) \(E \alpha \dfrac{1}{m}\) As mass for deuteron is maximum so its energy is minimum.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357932
A proton and an electron have the same de-Broglie wavelength. If \(K_{p}\) and \(K_{e}\) be the kinetic energies of proton and electron respectively, then choose the correct relation.
1 \(K_{p}=K_{e}\)
2 \(K_{p}>K_{e}\)
3 \(K_{p}=K_{e}^{2}\)
4 \(K_{p} < K_{e}\)
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\). If \(\lambda\) is same, so momentum \(p\) will be same. Now, kinetice energy \(K=\dfrac{p^{2}}{2 m}\) \(\therefore K \propto \dfrac{1}{m} ; m_{p}>m_{e} \Rightarrow K_{p} < K_{e}\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357933
Electron behaves as wave because they can be
1 deflected by an electric field
2 deflected by magnetic field
3 diffracted by a crystal
4 they ionize a gas
Explanation:
Electrons show the diffraction phenomenon therefore they can behave as a wave also. So correct option is (3)
357931
Which of the following particles - neutron, proton, electron and deuteron has the lowest energy if all have the same de Broglie wavelength?
1 Neutron
2 Proton
3 Electron
4 Deuteron
Explanation:
\(\lambda=\dfrac{h}{\sqrt{2 m E}} \Rightarrow E=\dfrac{h^{2}}{2 m \lambda^{2}}\) \(E \alpha \dfrac{1}{m}\) As mass for deuteron is maximum so its energy is minimum.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357932
A proton and an electron have the same de-Broglie wavelength. If \(K_{p}\) and \(K_{e}\) be the kinetic energies of proton and electron respectively, then choose the correct relation.
1 \(K_{p}=K_{e}\)
2 \(K_{p}>K_{e}\)
3 \(K_{p}=K_{e}^{2}\)
4 \(K_{p} < K_{e}\)
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\). If \(\lambda\) is same, so momentum \(p\) will be same. Now, kinetice energy \(K=\dfrac{p^{2}}{2 m}\) \(\therefore K \propto \dfrac{1}{m} ; m_{p}>m_{e} \Rightarrow K_{p} < K_{e}\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357933
Electron behaves as wave because they can be
1 deflected by an electric field
2 deflected by magnetic field
3 diffracted by a crystal
4 they ionize a gas
Explanation:
Electrons show the diffraction phenomenon therefore they can behave as a wave also. So correct option is (3)
357931
Which of the following particles - neutron, proton, electron and deuteron has the lowest energy if all have the same de Broglie wavelength?
1 Neutron
2 Proton
3 Electron
4 Deuteron
Explanation:
\(\lambda=\dfrac{h}{\sqrt{2 m E}} \Rightarrow E=\dfrac{h^{2}}{2 m \lambda^{2}}\) \(E \alpha \dfrac{1}{m}\) As mass for deuteron is maximum so its energy is minimum.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357932
A proton and an electron have the same de-Broglie wavelength. If \(K_{p}\) and \(K_{e}\) be the kinetic energies of proton and electron respectively, then choose the correct relation.
1 \(K_{p}=K_{e}\)
2 \(K_{p}>K_{e}\)
3 \(K_{p}=K_{e}^{2}\)
4 \(K_{p} < K_{e}\)
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\). If \(\lambda\) is same, so momentum \(p\) will be same. Now, kinetice energy \(K=\dfrac{p^{2}}{2 m}\) \(\therefore K \propto \dfrac{1}{m} ; m_{p}>m_{e} \Rightarrow K_{p} < K_{e}\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357933
Electron behaves as wave because they can be
1 deflected by an electric field
2 deflected by magnetic field
3 diffracted by a crystal
4 they ionize a gas
Explanation:
Electrons show the diffraction phenomenon therefore they can behave as a wave also. So correct option is (3)
357931
Which of the following particles - neutron, proton, electron and deuteron has the lowest energy if all have the same de Broglie wavelength?
1 Neutron
2 Proton
3 Electron
4 Deuteron
Explanation:
\(\lambda=\dfrac{h}{\sqrt{2 m E}} \Rightarrow E=\dfrac{h^{2}}{2 m \lambda^{2}}\) \(E \alpha \dfrac{1}{m}\) As mass for deuteron is maximum so its energy is minimum.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357932
A proton and an electron have the same de-Broglie wavelength. If \(K_{p}\) and \(K_{e}\) be the kinetic energies of proton and electron respectively, then choose the correct relation.
1 \(K_{p}=K_{e}\)
2 \(K_{p}>K_{e}\)
3 \(K_{p}=K_{e}^{2}\)
4 \(K_{p} < K_{e}\)
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\). If \(\lambda\) is same, so momentum \(p\) will be same. Now, kinetice energy \(K=\dfrac{p^{2}}{2 m}\) \(\therefore K \propto \dfrac{1}{m} ; m_{p}>m_{e} \Rightarrow K_{p} < K_{e}\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357933
Electron behaves as wave because they can be
1 deflected by an electric field
2 deflected by magnetic field
3 diffracted by a crystal
4 they ionize a gas
Explanation:
Electrons show the diffraction phenomenon therefore they can behave as a wave also. So correct option is (3)
