357914
An electron is accelerated through a potential difference of 10,000 \(V\). Its de Broglie wavelength is, (nearly) : \(\left( {{m_e} = 9 \times {{10}^{ - 31}}\;kg} \right)\)
1 \(12.2 \times {10^{ - 13}}\;m\)
2 \(12.2 \times {10^{ - 12}}\;m\)
3 \(12.2 \times {10^{ - 14}}\;m\)
4 \(12.2\;nm\)
Explanation:
For an electron accelerated through a potential \(V\) \(\lambda = \frac{{12.27}}{{\sqrt V }}\mathop A\limits^o = \frac{{12.27 \times {{10}^{ - 10}}}}{{\sqrt {10000} }} = 12.27 \times {10^{ - 12}}\;m\)
NEET - 2019
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357915
A particle of mass \(M\) at rest decays into two particles of masses \(m_{1}\) and \(m_{2}\), having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, \(\lambda_{1} / \lambda_{2}\), is
1 \(m_{2} / m_{1}\)
2 \(m_{1} / m_{2}\)
3 \(\sqrt{m_{2}} / \sqrt{m_{1}}\)
4 1.0
Explanation:
de-Broglie wavelength \(\lambda=\dfrac{h}{p} \therefore \dfrac{\lambda_{1}}{\lambda_{2}}=\dfrac{p_{2}}{p_{1}}\) And momentum is conserved in decay so \(p_{1}=p_{2}\) and \(\dfrac{\lambda_{1}}{\lambda_{2}}=1\).
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357916
The de-Broglie wavelength of an electron having kinetic energy \(E\) is \(\lambda\). If the kinetic energy of electron becomes \(\dfrac{E}{4}\), then its de-Broglie wavelength will be
1 \(\dfrac{\lambda}{\sqrt{2}}\)
2 \(2 \lambda\)
3 \(\sqrt{2} \lambda\)
4 \(\dfrac{\lambda}{2}\)
Explanation:
If \(p\) be the linear momentum of a moving particle and \(m\) be its mass, then its kinetic energy is given by \(E=p^{2} / 2 m\) \(\therefore p=\sqrt{2 m E}\). Therefore de Broglie equation is given by \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E}} \Rightarrow \lambda \dfrac{1}{\sqrt{E}}\). \(\therefore \dfrac{\lambda^{\prime}}{\lambda}=\sqrt{\dfrac{E}{E^{\prime}}}=\sqrt{\dfrac{E}{E / 4}}=2 \Rightarrow \lambda^{\prime}=2 \lambda\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357917
For a given kinetic energy which of the following has the smallest de-Broglie wavelength?
1 Electron
2 Proton
3 Neutron
4 \(\alpha\)-particle
Explanation:
De-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For a given kinetic energy \(\lambda \alpha \dfrac{1}{\sqrt{m}}\) \(\alpha\)-particle has largest mass. So it has smallest de-Broglie wavelength.
357914
An electron is accelerated through a potential difference of 10,000 \(V\). Its de Broglie wavelength is, (nearly) : \(\left( {{m_e} = 9 \times {{10}^{ - 31}}\;kg} \right)\)
1 \(12.2 \times {10^{ - 13}}\;m\)
2 \(12.2 \times {10^{ - 12}}\;m\)
3 \(12.2 \times {10^{ - 14}}\;m\)
4 \(12.2\;nm\)
Explanation:
For an electron accelerated through a potential \(V\) \(\lambda = \frac{{12.27}}{{\sqrt V }}\mathop A\limits^o = \frac{{12.27 \times {{10}^{ - 10}}}}{{\sqrt {10000} }} = 12.27 \times {10^{ - 12}}\;m\)
NEET - 2019
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357915
A particle of mass \(M\) at rest decays into two particles of masses \(m_{1}\) and \(m_{2}\), having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, \(\lambda_{1} / \lambda_{2}\), is
1 \(m_{2} / m_{1}\)
2 \(m_{1} / m_{2}\)
3 \(\sqrt{m_{2}} / \sqrt{m_{1}}\)
4 1.0
Explanation:
de-Broglie wavelength \(\lambda=\dfrac{h}{p} \therefore \dfrac{\lambda_{1}}{\lambda_{2}}=\dfrac{p_{2}}{p_{1}}\) And momentum is conserved in decay so \(p_{1}=p_{2}\) and \(\dfrac{\lambda_{1}}{\lambda_{2}}=1\).
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357916
The de-Broglie wavelength of an electron having kinetic energy \(E\) is \(\lambda\). If the kinetic energy of electron becomes \(\dfrac{E}{4}\), then its de-Broglie wavelength will be
1 \(\dfrac{\lambda}{\sqrt{2}}\)
2 \(2 \lambda\)
3 \(\sqrt{2} \lambda\)
4 \(\dfrac{\lambda}{2}\)
Explanation:
If \(p\) be the linear momentum of a moving particle and \(m\) be its mass, then its kinetic energy is given by \(E=p^{2} / 2 m\) \(\therefore p=\sqrt{2 m E}\). Therefore de Broglie equation is given by \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E}} \Rightarrow \lambda \dfrac{1}{\sqrt{E}}\). \(\therefore \dfrac{\lambda^{\prime}}{\lambda}=\sqrt{\dfrac{E}{E^{\prime}}}=\sqrt{\dfrac{E}{E / 4}}=2 \Rightarrow \lambda^{\prime}=2 \lambda\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357917
For a given kinetic energy which of the following has the smallest de-Broglie wavelength?
1 Electron
2 Proton
3 Neutron
4 \(\alpha\)-particle
Explanation:
De-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For a given kinetic energy \(\lambda \alpha \dfrac{1}{\sqrt{m}}\) \(\alpha\)-particle has largest mass. So it has smallest de-Broglie wavelength.
357914
An electron is accelerated through a potential difference of 10,000 \(V\). Its de Broglie wavelength is, (nearly) : \(\left( {{m_e} = 9 \times {{10}^{ - 31}}\;kg} \right)\)
1 \(12.2 \times {10^{ - 13}}\;m\)
2 \(12.2 \times {10^{ - 12}}\;m\)
3 \(12.2 \times {10^{ - 14}}\;m\)
4 \(12.2\;nm\)
Explanation:
For an electron accelerated through a potential \(V\) \(\lambda = \frac{{12.27}}{{\sqrt V }}\mathop A\limits^o = \frac{{12.27 \times {{10}^{ - 10}}}}{{\sqrt {10000} }} = 12.27 \times {10^{ - 12}}\;m\)
NEET - 2019
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357915
A particle of mass \(M\) at rest decays into two particles of masses \(m_{1}\) and \(m_{2}\), having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, \(\lambda_{1} / \lambda_{2}\), is
1 \(m_{2} / m_{1}\)
2 \(m_{1} / m_{2}\)
3 \(\sqrt{m_{2}} / \sqrt{m_{1}}\)
4 1.0
Explanation:
de-Broglie wavelength \(\lambda=\dfrac{h}{p} \therefore \dfrac{\lambda_{1}}{\lambda_{2}}=\dfrac{p_{2}}{p_{1}}\) And momentum is conserved in decay so \(p_{1}=p_{2}\) and \(\dfrac{\lambda_{1}}{\lambda_{2}}=1\).
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357916
The de-Broglie wavelength of an electron having kinetic energy \(E\) is \(\lambda\). If the kinetic energy of electron becomes \(\dfrac{E}{4}\), then its de-Broglie wavelength will be
1 \(\dfrac{\lambda}{\sqrt{2}}\)
2 \(2 \lambda\)
3 \(\sqrt{2} \lambda\)
4 \(\dfrac{\lambda}{2}\)
Explanation:
If \(p\) be the linear momentum of a moving particle and \(m\) be its mass, then its kinetic energy is given by \(E=p^{2} / 2 m\) \(\therefore p=\sqrt{2 m E}\). Therefore de Broglie equation is given by \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E}} \Rightarrow \lambda \dfrac{1}{\sqrt{E}}\). \(\therefore \dfrac{\lambda^{\prime}}{\lambda}=\sqrt{\dfrac{E}{E^{\prime}}}=\sqrt{\dfrac{E}{E / 4}}=2 \Rightarrow \lambda^{\prime}=2 \lambda\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357917
For a given kinetic energy which of the following has the smallest de-Broglie wavelength?
1 Electron
2 Proton
3 Neutron
4 \(\alpha\)-particle
Explanation:
De-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For a given kinetic energy \(\lambda \alpha \dfrac{1}{\sqrt{m}}\) \(\alpha\)-particle has largest mass. So it has smallest de-Broglie wavelength.
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PHXII11:DUAL NATURE OF RADIATION AND MATTER
357914
An electron is accelerated through a potential difference of 10,000 \(V\). Its de Broglie wavelength is, (nearly) : \(\left( {{m_e} = 9 \times {{10}^{ - 31}}\;kg} \right)\)
1 \(12.2 \times {10^{ - 13}}\;m\)
2 \(12.2 \times {10^{ - 12}}\;m\)
3 \(12.2 \times {10^{ - 14}}\;m\)
4 \(12.2\;nm\)
Explanation:
For an electron accelerated through a potential \(V\) \(\lambda = \frac{{12.27}}{{\sqrt V }}\mathop A\limits^o = \frac{{12.27 \times {{10}^{ - 10}}}}{{\sqrt {10000} }} = 12.27 \times {10^{ - 12}}\;m\)
NEET - 2019
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357915
A particle of mass \(M\) at rest decays into two particles of masses \(m_{1}\) and \(m_{2}\), having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, \(\lambda_{1} / \lambda_{2}\), is
1 \(m_{2} / m_{1}\)
2 \(m_{1} / m_{2}\)
3 \(\sqrt{m_{2}} / \sqrt{m_{1}}\)
4 1.0
Explanation:
de-Broglie wavelength \(\lambda=\dfrac{h}{p} \therefore \dfrac{\lambda_{1}}{\lambda_{2}}=\dfrac{p_{2}}{p_{1}}\) And momentum is conserved in decay so \(p_{1}=p_{2}\) and \(\dfrac{\lambda_{1}}{\lambda_{2}}=1\).
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357916
The de-Broglie wavelength of an electron having kinetic energy \(E\) is \(\lambda\). If the kinetic energy of electron becomes \(\dfrac{E}{4}\), then its de-Broglie wavelength will be
1 \(\dfrac{\lambda}{\sqrt{2}}\)
2 \(2 \lambda\)
3 \(\sqrt{2} \lambda\)
4 \(\dfrac{\lambda}{2}\)
Explanation:
If \(p\) be the linear momentum of a moving particle and \(m\) be its mass, then its kinetic energy is given by \(E=p^{2} / 2 m\) \(\therefore p=\sqrt{2 m E}\). Therefore de Broglie equation is given by \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E}} \Rightarrow \lambda \dfrac{1}{\sqrt{E}}\). \(\therefore \dfrac{\lambda^{\prime}}{\lambda}=\sqrt{\dfrac{E}{E^{\prime}}}=\sqrt{\dfrac{E}{E / 4}}=2 \Rightarrow \lambda^{\prime}=2 \lambda\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357917
For a given kinetic energy which of the following has the smallest de-Broglie wavelength?
1 Electron
2 Proton
3 Neutron
4 \(\alpha\)-particle
Explanation:
De-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For a given kinetic energy \(\lambda \alpha \dfrac{1}{\sqrt{m}}\) \(\alpha\)-particle has largest mass. So it has smallest de-Broglie wavelength.
