NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142479
The log-log graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be
1
2
3
4
Explanation:
A We know that, de-Broglie wavelength - $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}}} \times \frac{1}{\sqrt{\mathrm{E}}}$ Taking $\log$ on both side $\log \lambda=\log \frac{h}{\sqrt{2} m}+\log \frac{1}{\sqrt{E}}$ $\log \lambda=\log \frac{h}{\sqrt{2} m}-\frac{1}{2} \log E$ $\log \lambda=-\frac{1}{2} \log E+\log \frac{h}{\sqrt{2} m}$ The equation of straight line having slope $(-1 / 2)$ and positive intercept on $\log \lambda$ axis.
UPSEE - 2015
Dual nature of radiation and Matter
142480
When the momentum of a proton is changed by an amount $p_{0}$, then the corresponding change in the de-Broglie wavelength is found to be $0.25 \%$. Then, the original momentum of the proton was
142481
A particle is dropped from a height $H$. The deBroglie wavelength of the particle as a function of height is proportional to
1 $\mathrm{H}$
2 $\mathrm{H}^{1 / 2}$
3 $\mathrm{H}^{0}$
4 $\mathrm{H}^{-1 / 2}$
Explanation:
D de - Broglie wavelength $\lambda =\frac{\mathrm{h}}{\mathrm{p}} \quad(\because \mathrm{p}=\mathrm{mv})$ $\lambda =\frac{\mathrm{h}}{\mathrm{mv}}$ Velocity acquired by a particle while falling from a height $\mathrm{H}, \mathrm{v}=\sqrt{2 \mathrm{gH}}$ Putting the value of $v$ from equation (ii) in equation (i) $\therefore \quad \lambda =\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{gH}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{H}}}$ $\lambda \propto \mathrm{H}^{-1 / 2}$
UPSEE - 2013
Dual nature of radiation and Matter
142484
The kinetic energy of an electron with de Broglie wavelength of $\mathbf{3} \mathbf{~ n m}$ is :
142479
The log-log graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be
1
2
3
4
Explanation:
A We know that, de-Broglie wavelength - $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}}} \times \frac{1}{\sqrt{\mathrm{E}}}$ Taking $\log$ on both side $\log \lambda=\log \frac{h}{\sqrt{2} m}+\log \frac{1}{\sqrt{E}}$ $\log \lambda=\log \frac{h}{\sqrt{2} m}-\frac{1}{2} \log E$ $\log \lambda=-\frac{1}{2} \log E+\log \frac{h}{\sqrt{2} m}$ The equation of straight line having slope $(-1 / 2)$ and positive intercept on $\log \lambda$ axis.
UPSEE - 2015
Dual nature of radiation and Matter
142480
When the momentum of a proton is changed by an amount $p_{0}$, then the corresponding change in the de-Broglie wavelength is found to be $0.25 \%$. Then, the original momentum of the proton was
142481
A particle is dropped from a height $H$. The deBroglie wavelength of the particle as a function of height is proportional to
1 $\mathrm{H}$
2 $\mathrm{H}^{1 / 2}$
3 $\mathrm{H}^{0}$
4 $\mathrm{H}^{-1 / 2}$
Explanation:
D de - Broglie wavelength $\lambda =\frac{\mathrm{h}}{\mathrm{p}} \quad(\because \mathrm{p}=\mathrm{mv})$ $\lambda =\frac{\mathrm{h}}{\mathrm{mv}}$ Velocity acquired by a particle while falling from a height $\mathrm{H}, \mathrm{v}=\sqrt{2 \mathrm{gH}}$ Putting the value of $v$ from equation (ii) in equation (i) $\therefore \quad \lambda =\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{gH}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{H}}}$ $\lambda \propto \mathrm{H}^{-1 / 2}$
UPSEE - 2013
Dual nature of radiation and Matter
142484
The kinetic energy of an electron with de Broglie wavelength of $\mathbf{3} \mathbf{~ n m}$ is :
142479
The log-log graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be
1
2
3
4
Explanation:
A We know that, de-Broglie wavelength - $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}}} \times \frac{1}{\sqrt{\mathrm{E}}}$ Taking $\log$ on both side $\log \lambda=\log \frac{h}{\sqrt{2} m}+\log \frac{1}{\sqrt{E}}$ $\log \lambda=\log \frac{h}{\sqrt{2} m}-\frac{1}{2} \log E$ $\log \lambda=-\frac{1}{2} \log E+\log \frac{h}{\sqrt{2} m}$ The equation of straight line having slope $(-1 / 2)$ and positive intercept on $\log \lambda$ axis.
UPSEE - 2015
Dual nature of radiation and Matter
142480
When the momentum of a proton is changed by an amount $p_{0}$, then the corresponding change in the de-Broglie wavelength is found to be $0.25 \%$. Then, the original momentum of the proton was
142481
A particle is dropped from a height $H$. The deBroglie wavelength of the particle as a function of height is proportional to
1 $\mathrm{H}$
2 $\mathrm{H}^{1 / 2}$
3 $\mathrm{H}^{0}$
4 $\mathrm{H}^{-1 / 2}$
Explanation:
D de - Broglie wavelength $\lambda =\frac{\mathrm{h}}{\mathrm{p}} \quad(\because \mathrm{p}=\mathrm{mv})$ $\lambda =\frac{\mathrm{h}}{\mathrm{mv}}$ Velocity acquired by a particle while falling from a height $\mathrm{H}, \mathrm{v}=\sqrt{2 \mathrm{gH}}$ Putting the value of $v$ from equation (ii) in equation (i) $\therefore \quad \lambda =\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{gH}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{H}}}$ $\lambda \propto \mathrm{H}^{-1 / 2}$
UPSEE - 2013
Dual nature of radiation and Matter
142484
The kinetic energy of an electron with de Broglie wavelength of $\mathbf{3} \mathbf{~ n m}$ is :
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Dual nature of radiation and Matter
142479
The log-log graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be
1
2
3
4
Explanation:
A We know that, de-Broglie wavelength - $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}}} \times \frac{1}{\sqrt{\mathrm{E}}}$ Taking $\log$ on both side $\log \lambda=\log \frac{h}{\sqrt{2} m}+\log \frac{1}{\sqrt{E}}$ $\log \lambda=\log \frac{h}{\sqrt{2} m}-\frac{1}{2} \log E$ $\log \lambda=-\frac{1}{2} \log E+\log \frac{h}{\sqrt{2} m}$ The equation of straight line having slope $(-1 / 2)$ and positive intercept on $\log \lambda$ axis.
UPSEE - 2015
Dual nature of radiation and Matter
142480
When the momentum of a proton is changed by an amount $p_{0}$, then the corresponding change in the de-Broglie wavelength is found to be $0.25 \%$. Then, the original momentum of the proton was
142481
A particle is dropped from a height $H$. The deBroglie wavelength of the particle as a function of height is proportional to
1 $\mathrm{H}$
2 $\mathrm{H}^{1 / 2}$
3 $\mathrm{H}^{0}$
4 $\mathrm{H}^{-1 / 2}$
Explanation:
D de - Broglie wavelength $\lambda =\frac{\mathrm{h}}{\mathrm{p}} \quad(\because \mathrm{p}=\mathrm{mv})$ $\lambda =\frac{\mathrm{h}}{\mathrm{mv}}$ Velocity acquired by a particle while falling from a height $\mathrm{H}, \mathrm{v}=\sqrt{2 \mathrm{gH}}$ Putting the value of $v$ from equation (ii) in equation (i) $\therefore \quad \lambda =\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{gH}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{H}}}$ $\lambda \propto \mathrm{H}^{-1 / 2}$
UPSEE - 2013
Dual nature of radiation and Matter
142484
The kinetic energy of an electron with de Broglie wavelength of $\mathbf{3} \mathbf{~ n m}$ is :
