Wave Nature Of Light Of Matter (de-Broglie)
Dual nature of radiation and Matter

142447 An electron on charge e and mass $m$ moving with an initial velocity $v_{0} \hat{i}$ is subjected to all electric field $E_{0} \hat{j}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron = $\left.\lambda_{0}\right)$

1 $\lambda_{0}$
2 $\lambda_{0} \sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}$
3 $\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}}$
4 $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{m v_{0}^{2}}\right)}$
Dual nature of radiation and Matter

142448 An electron of mass $m$ with a velocity $v=$ $v_{0} \hat{i}\left(v_{0}>0\right)$ enters an electric field $E=-E_{0} \hat{i}\left(E_{0}\right.$ constant $>0$ ) at $t=0$. If $\lambda_{0}$ its de-Broglie wavelength initially, then its de-Broglie wavelength at time is

1 $\lambda_{0} \mathrm{t}$
2 $\lambda_{0}\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mv}_{0}} \mathrm{t}\right)$
3 $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mv}_{0}} \mathrm{t}\right)}$
4 $\lambda_{0}$
Dual nature of radiation and Matter

142450 An electron accelerated through a potential of $10000 \mathrm{~V}$ from rest has a de-Broglie wave length ' $\lambda$ '. What should be the accelerating potential, so that the wavelength is doubled?

1 $20000 \mathrm{~V}$
2 $40000 \mathrm{~V}$
3 $5000 \mathrm{~V}$
4 $2500 \mathrm{~V}$
Dual nature of radiation and Matter

142451 If an electron and a proton have the same deBroglie wavelength, then the kinetic energy of the electron is :

1 zero
2 less than that of a proton
3 more than that of a proton
4 equal to that of a proton
Dual nature of radiation and Matter

142447 An electron on charge e and mass $m$ moving with an initial velocity $v_{0} \hat{i}$ is subjected to all electric field $E_{0} \hat{j}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron = $\left.\lambda_{0}\right)$

1 $\lambda_{0}$
2 $\lambda_{0} \sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}$
3 $\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}}$
4 $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{m v_{0}^{2}}\right)}$
Dual nature of radiation and Matter

142448 An electron of mass $m$ with a velocity $v=$ $v_{0} \hat{i}\left(v_{0}>0\right)$ enters an electric field $E=-E_{0} \hat{i}\left(E_{0}\right.$ constant $>0$ ) at $t=0$. If $\lambda_{0}$ its de-Broglie wavelength initially, then its de-Broglie wavelength at time is

1 $\lambda_{0} \mathrm{t}$
2 $\lambda_{0}\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mv}_{0}} \mathrm{t}\right)$
3 $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mv}_{0}} \mathrm{t}\right)}$
4 $\lambda_{0}$
Dual nature of radiation and Matter

142450 An electron accelerated through a potential of $10000 \mathrm{~V}$ from rest has a de-Broglie wave length ' $\lambda$ '. What should be the accelerating potential, so that the wavelength is doubled?

1 $20000 \mathrm{~V}$
2 $40000 \mathrm{~V}$
3 $5000 \mathrm{~V}$
4 $2500 \mathrm{~V}$
Dual nature of radiation and Matter

142451 If an electron and a proton have the same deBroglie wavelength, then the kinetic energy of the electron is :

1 zero
2 less than that of a proton
3 more than that of a proton
4 equal to that of a proton
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Dual nature of radiation and Matter

142447 An electron on charge e and mass $m$ moving with an initial velocity $v_{0} \hat{i}$ is subjected to all electric field $E_{0} \hat{j}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron = $\left.\lambda_{0}\right)$

1 $\lambda_{0}$
2 $\lambda_{0} \sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}$
3 $\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}}$
4 $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{m v_{0}^{2}}\right)}$
Dual nature of radiation and Matter

142448 An electron of mass $m$ with a velocity $v=$ $v_{0} \hat{i}\left(v_{0}>0\right)$ enters an electric field $E=-E_{0} \hat{i}\left(E_{0}\right.$ constant $>0$ ) at $t=0$. If $\lambda_{0}$ its de-Broglie wavelength initially, then its de-Broglie wavelength at time is

1 $\lambda_{0} \mathrm{t}$
2 $\lambda_{0}\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mv}_{0}} \mathrm{t}\right)$
3 $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mv}_{0}} \mathrm{t}\right)}$
4 $\lambda_{0}$
Dual nature of radiation and Matter

142450 An electron accelerated through a potential of $10000 \mathrm{~V}$ from rest has a de-Broglie wave length ' $\lambda$ '. What should be the accelerating potential, so that the wavelength is doubled?

1 $20000 \mathrm{~V}$
2 $40000 \mathrm{~V}$
3 $5000 \mathrm{~V}$
4 $2500 \mathrm{~V}$
Dual nature of radiation and Matter

142451 If an electron and a proton have the same deBroglie wavelength, then the kinetic energy of the electron is :

1 zero
2 less than that of a proton
3 more than that of a proton
4 equal to that of a proton
Dual nature of radiation and Matter

142447 An electron on charge e and mass $m$ moving with an initial velocity $v_{0} \hat{i}$ is subjected to all electric field $E_{0} \hat{j}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron = $\left.\lambda_{0}\right)$

1 $\lambda_{0}$
2 $\lambda_{0} \sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}$
3 $\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}}$
4 $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{m v_{0}^{2}}\right)}$
Dual nature of radiation and Matter

142448 An electron of mass $m$ with a velocity $v=$ $v_{0} \hat{i}\left(v_{0}>0\right)$ enters an electric field $E=-E_{0} \hat{i}\left(E_{0}\right.$ constant $>0$ ) at $t=0$. If $\lambda_{0}$ its de-Broglie wavelength initially, then its de-Broglie wavelength at time is

1 $\lambda_{0} \mathrm{t}$
2 $\lambda_{0}\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mv}_{0}} \mathrm{t}\right)$
3 $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mv}_{0}} \mathrm{t}\right)}$
4 $\lambda_{0}$
Dual nature of radiation and Matter

142450 An electron accelerated through a potential of $10000 \mathrm{~V}$ from rest has a de-Broglie wave length ' $\lambda$ '. What should be the accelerating potential, so that the wavelength is doubled?

1 $20000 \mathrm{~V}$
2 $40000 \mathrm{~V}$
3 $5000 \mathrm{~V}$
4 $2500 \mathrm{~V}$
Dual nature of radiation and Matter

142451 If an electron and a proton have the same deBroglie wavelength, then the kinetic energy of the electron is :

1 zero
2 less than that of a proton
3 more than that of a proton
4 equal to that of a proton