357879
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is \(25 \%\) of the velocity of light, then the ratio of \(K.E.\) of electron and \(K.E.\) of photon will be
1 \(\dfrac{1}{8}\)
2 \(\dfrac{1}{1}\)
3 \(\dfrac{8}{1}\)
4 \(\dfrac{1}{4}\)
Explanation:
\(\lambda_{e}=\lambda_{\text {photon }}\) Velocity of electron \(=25 \%\) velocity of light Kinetic energy of electron \(=E_{k}=\dfrac{1}{2} p v_{e}=\dfrac{h v_{e}}{2 \lambda_{e}}\left[\because p=\dfrac{h}{\lambda}\right]\) and \(K . E\). of photon \(=\dfrac{h c}{\lambda_{p}}\) \(\therefore\) Ratio \(=\dfrac{\text { K.E. of electron }}{\text { K.E. of photon }}=\dfrac{h v_{e}}{2 \lambda_{e}} \dfrac{\lambda_{p}}{h c}\) (Given, \(\lambda_{e}=\lambda_{p}\) ) \( = \frac{{{v_e}}}{{2\,c}} = \frac{{0.25\,c}}{{2\,c}} = \frac{{25}}{{200}} = \frac{1}{8}\quad \left( {{v_e} = 0.25\,c} \right)\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357880
If the momentum of electron is changed by \(P\), then the de Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be
357883
An electron is moving with an initial velocity \(v = {v_0}\widehat i\) in a magnetic field \(B = {B_0}j\). Then it's de Broglie wavelength
1 Increases with time
2 Remains constant
3 Increases and decreases periodically
4 Decreases with time
Explanation:
Here, when a particle is moving in a magnetic field then its speed remains constant. So the magnitude of \(v\) will not change, i.e., momentum ( \(mv\) ) will remain constant in magnitude. Hence de Broglie wavelength \(\lambda = h/mv\) remains constant.
357879
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is \(25 \%\) of the velocity of light, then the ratio of \(K.E.\) of electron and \(K.E.\) of photon will be
1 \(\dfrac{1}{8}\)
2 \(\dfrac{1}{1}\)
3 \(\dfrac{8}{1}\)
4 \(\dfrac{1}{4}\)
Explanation:
\(\lambda_{e}=\lambda_{\text {photon }}\) Velocity of electron \(=25 \%\) velocity of light Kinetic energy of electron \(=E_{k}=\dfrac{1}{2} p v_{e}=\dfrac{h v_{e}}{2 \lambda_{e}}\left[\because p=\dfrac{h}{\lambda}\right]\) and \(K . E\). of photon \(=\dfrac{h c}{\lambda_{p}}\) \(\therefore\) Ratio \(=\dfrac{\text { K.E. of electron }}{\text { K.E. of photon }}=\dfrac{h v_{e}}{2 \lambda_{e}} \dfrac{\lambda_{p}}{h c}\) (Given, \(\lambda_{e}=\lambda_{p}\) ) \( = \frac{{{v_e}}}{{2\,c}} = \frac{{0.25\,c}}{{2\,c}} = \frac{{25}}{{200}} = \frac{1}{8}\quad \left( {{v_e} = 0.25\,c} \right)\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357880
If the momentum of electron is changed by \(P\), then the de Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be
357883
An electron is moving with an initial velocity \(v = {v_0}\widehat i\) in a magnetic field \(B = {B_0}j\). Then it's de Broglie wavelength
1 Increases with time
2 Remains constant
3 Increases and decreases periodically
4 Decreases with time
Explanation:
Here, when a particle is moving in a magnetic field then its speed remains constant. So the magnitude of \(v\) will not change, i.e., momentum ( \(mv\) ) will remain constant in magnitude. Hence de Broglie wavelength \(\lambda = h/mv\) remains constant.
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PHXII11:DUAL NATURE OF RADIATION AND MATTER
357879
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is \(25 \%\) of the velocity of light, then the ratio of \(K.E.\) of electron and \(K.E.\) of photon will be
1 \(\dfrac{1}{8}\)
2 \(\dfrac{1}{1}\)
3 \(\dfrac{8}{1}\)
4 \(\dfrac{1}{4}\)
Explanation:
\(\lambda_{e}=\lambda_{\text {photon }}\) Velocity of electron \(=25 \%\) velocity of light Kinetic energy of electron \(=E_{k}=\dfrac{1}{2} p v_{e}=\dfrac{h v_{e}}{2 \lambda_{e}}\left[\because p=\dfrac{h}{\lambda}\right]\) and \(K . E\). of photon \(=\dfrac{h c}{\lambda_{p}}\) \(\therefore\) Ratio \(=\dfrac{\text { K.E. of electron }}{\text { K.E. of photon }}=\dfrac{h v_{e}}{2 \lambda_{e}} \dfrac{\lambda_{p}}{h c}\) (Given, \(\lambda_{e}=\lambda_{p}\) ) \( = \frac{{{v_e}}}{{2\,c}} = \frac{{0.25\,c}}{{2\,c}} = \frac{{25}}{{200}} = \frac{1}{8}\quad \left( {{v_e} = 0.25\,c} \right)\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357880
If the momentum of electron is changed by \(P\), then the de Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be
357883
An electron is moving with an initial velocity \(v = {v_0}\widehat i\) in a magnetic field \(B = {B_0}j\). Then it's de Broglie wavelength
1 Increases with time
2 Remains constant
3 Increases and decreases periodically
4 Decreases with time
Explanation:
Here, when a particle is moving in a magnetic field then its speed remains constant. So the magnitude of \(v\) will not change, i.e., momentum ( \(mv\) ) will remain constant in magnitude. Hence de Broglie wavelength \(\lambda = h/mv\) remains constant.
357879
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is \(25 \%\) of the velocity of light, then the ratio of \(K.E.\) of electron and \(K.E.\) of photon will be
1 \(\dfrac{1}{8}\)
2 \(\dfrac{1}{1}\)
3 \(\dfrac{8}{1}\)
4 \(\dfrac{1}{4}\)
Explanation:
\(\lambda_{e}=\lambda_{\text {photon }}\) Velocity of electron \(=25 \%\) velocity of light Kinetic energy of electron \(=E_{k}=\dfrac{1}{2} p v_{e}=\dfrac{h v_{e}}{2 \lambda_{e}}\left[\because p=\dfrac{h}{\lambda}\right]\) and \(K . E\). of photon \(=\dfrac{h c}{\lambda_{p}}\) \(\therefore\) Ratio \(=\dfrac{\text { K.E. of electron }}{\text { K.E. of photon }}=\dfrac{h v_{e}}{2 \lambda_{e}} \dfrac{\lambda_{p}}{h c}\) (Given, \(\lambda_{e}=\lambda_{p}\) ) \( = \frac{{{v_e}}}{{2\,c}} = \frac{{0.25\,c}}{{2\,c}} = \frac{{25}}{{200}} = \frac{1}{8}\quad \left( {{v_e} = 0.25\,c} \right)\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357880
If the momentum of electron is changed by \(P\), then the de Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be
357883
An electron is moving with an initial velocity \(v = {v_0}\widehat i\) in a magnetic field \(B = {B_0}j\). Then it's de Broglie wavelength
1 Increases with time
2 Remains constant
3 Increases and decreases periodically
4 Decreases with time
Explanation:
Here, when a particle is moving in a magnetic field then its speed remains constant. So the magnitude of \(v\) will not change, i.e., momentum ( \(mv\) ) will remain constant in magnitude. Hence de Broglie wavelength \(\lambda = h/mv\) remains constant.
357879
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is \(25 \%\) of the velocity of light, then the ratio of \(K.E.\) of electron and \(K.E.\) of photon will be
1 \(\dfrac{1}{8}\)
2 \(\dfrac{1}{1}\)
3 \(\dfrac{8}{1}\)
4 \(\dfrac{1}{4}\)
Explanation:
\(\lambda_{e}=\lambda_{\text {photon }}\) Velocity of electron \(=25 \%\) velocity of light Kinetic energy of electron \(=E_{k}=\dfrac{1}{2} p v_{e}=\dfrac{h v_{e}}{2 \lambda_{e}}\left[\because p=\dfrac{h}{\lambda}\right]\) and \(K . E\). of photon \(=\dfrac{h c}{\lambda_{p}}\) \(\therefore\) Ratio \(=\dfrac{\text { K.E. of electron }}{\text { K.E. of photon }}=\dfrac{h v_{e}}{2 \lambda_{e}} \dfrac{\lambda_{p}}{h c}\) (Given, \(\lambda_{e}=\lambda_{p}\) ) \( = \frac{{{v_e}}}{{2\,c}} = \frac{{0.25\,c}}{{2\,c}} = \frac{{25}}{{200}} = \frac{1}{8}\quad \left( {{v_e} = 0.25\,c} \right)\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357880
If the momentum of electron is changed by \(P\), then the de Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be
357883
An electron is moving with an initial velocity \(v = {v_0}\widehat i\) in a magnetic field \(B = {B_0}j\). Then it's de Broglie wavelength
1 Increases with time
2 Remains constant
3 Increases and decreases periodically
4 Decreases with time
Explanation:
Here, when a particle is moving in a magnetic field then its speed remains constant. So the magnitude of \(v\) will not change, i.e., momentum ( \(mv\) ) will remain constant in magnitude. Hence de Broglie wavelength \(\lambda = h/mv\) remains constant.