142505
If the kinetic energy of free electron is made double, the new de-Broglie wave length will be times that of initial wave length.
1 $\sqrt{2}$
2 $\frac{1}{\sqrt{2}}$
3 2
4 $\frac{1}{2}$
Explanation:
B We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Kinetic energy of an moving electron, $\mathrm{K} =\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}^{2} =\frac{2 \mathrm{~K}}{\mathrm{~m}}$ Put the above value of $v$ in equation (i) we get, $\lambda=\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Case first:- $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}}}$ For second case:- $\mathrm{K}_{2}=2 \mathrm{~K}_{1}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{2}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times 2 \mathrm{~K}_{1}}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}} \times \sqrt{2}}$ Substituting $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}}}$ in the equation of $\lambda_{2}$ $\lambda_{2}=\frac{1}{\sqrt{2}} \lambda_{1}$ So, new wavelength is $\frac{1}{\sqrt{2}}$ times of the initial wavelength.
GUJCET 2014
Dual nature of radiation and Matter
142506
The de-Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is
1 $\frac{3.08}{\sqrt{\mathrm{T}}} \AA$
2 $\frac{0.308}{\sqrt{\mathrm{T}}} \AA$
3 $\frac{0.0308}{\sqrt{\mathrm{T}}} \AA$
4 $\frac{30.8}{\sqrt{\mathrm{T}}} \AA$
Explanation:
D Given that, $\mathrm{m}=1.67 \times 10^{-27} \mathrm{~kg}, \mathrm{~K}_{\mathrm{B}}=1.38 \times$ $10^{-23}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ We know that, de-Broglie wavelength of neutron at thermal equilibrium, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{\mathrm{B}} \mathrm{T}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times \mathrm{T}}}$ $\lambda=\frac{30.8 \times 10^{-10}}{\sqrt{\mathrm{T}} \mathrm{m}}$ $\lambda=\frac{30.8}{\sqrt{\mathrm{T}}} \AA$
COMEDK 2013
Dual nature of radiation and Matter
142507
The de-Broglie wavelength of an electron, $\alpha$ particle and a proton all having the same kinetic energy is respectively given as $\lambda_{\mathrm{e}}, \lambda_{\alpha}$ and $\lambda_{\mathrm{p}}$. Then which of the following is not true?
142505
If the kinetic energy of free electron is made double, the new de-Broglie wave length will be times that of initial wave length.
1 $\sqrt{2}$
2 $\frac{1}{\sqrt{2}}$
3 2
4 $\frac{1}{2}$
Explanation:
B We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Kinetic energy of an moving electron, $\mathrm{K} =\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}^{2} =\frac{2 \mathrm{~K}}{\mathrm{~m}}$ Put the above value of $v$ in equation (i) we get, $\lambda=\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Case first:- $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}}}$ For second case:- $\mathrm{K}_{2}=2 \mathrm{~K}_{1}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{2}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times 2 \mathrm{~K}_{1}}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}} \times \sqrt{2}}$ Substituting $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}}}$ in the equation of $\lambda_{2}$ $\lambda_{2}=\frac{1}{\sqrt{2}} \lambda_{1}$ So, new wavelength is $\frac{1}{\sqrt{2}}$ times of the initial wavelength.
GUJCET 2014
Dual nature of radiation and Matter
142506
The de-Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is
1 $\frac{3.08}{\sqrt{\mathrm{T}}} \AA$
2 $\frac{0.308}{\sqrt{\mathrm{T}}} \AA$
3 $\frac{0.0308}{\sqrt{\mathrm{T}}} \AA$
4 $\frac{30.8}{\sqrt{\mathrm{T}}} \AA$
Explanation:
D Given that, $\mathrm{m}=1.67 \times 10^{-27} \mathrm{~kg}, \mathrm{~K}_{\mathrm{B}}=1.38 \times$ $10^{-23}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ We know that, de-Broglie wavelength of neutron at thermal equilibrium, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{\mathrm{B}} \mathrm{T}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times \mathrm{T}}}$ $\lambda=\frac{30.8 \times 10^{-10}}{\sqrt{\mathrm{T}} \mathrm{m}}$ $\lambda=\frac{30.8}{\sqrt{\mathrm{T}}} \AA$
COMEDK 2013
Dual nature of radiation and Matter
142507
The de-Broglie wavelength of an electron, $\alpha$ particle and a proton all having the same kinetic energy is respectively given as $\lambda_{\mathrm{e}}, \lambda_{\alpha}$ and $\lambda_{\mathrm{p}}$. Then which of the following is not true?
142505
If the kinetic energy of free electron is made double, the new de-Broglie wave length will be times that of initial wave length.
1 $\sqrt{2}$
2 $\frac{1}{\sqrt{2}}$
3 2
4 $\frac{1}{2}$
Explanation:
B We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Kinetic energy of an moving electron, $\mathrm{K} =\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}^{2} =\frac{2 \mathrm{~K}}{\mathrm{~m}}$ Put the above value of $v$ in equation (i) we get, $\lambda=\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Case first:- $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}}}$ For second case:- $\mathrm{K}_{2}=2 \mathrm{~K}_{1}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{2}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times 2 \mathrm{~K}_{1}}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}} \times \sqrt{2}}$ Substituting $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}}}$ in the equation of $\lambda_{2}$ $\lambda_{2}=\frac{1}{\sqrt{2}} \lambda_{1}$ So, new wavelength is $\frac{1}{\sqrt{2}}$ times of the initial wavelength.
GUJCET 2014
Dual nature of radiation and Matter
142506
The de-Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is
1 $\frac{3.08}{\sqrt{\mathrm{T}}} \AA$
2 $\frac{0.308}{\sqrt{\mathrm{T}}} \AA$
3 $\frac{0.0308}{\sqrt{\mathrm{T}}} \AA$
4 $\frac{30.8}{\sqrt{\mathrm{T}}} \AA$
Explanation:
D Given that, $\mathrm{m}=1.67 \times 10^{-27} \mathrm{~kg}, \mathrm{~K}_{\mathrm{B}}=1.38 \times$ $10^{-23}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ We know that, de-Broglie wavelength of neutron at thermal equilibrium, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{\mathrm{B}} \mathrm{T}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times \mathrm{T}}}$ $\lambda=\frac{30.8 \times 10^{-10}}{\sqrt{\mathrm{T}} \mathrm{m}}$ $\lambda=\frac{30.8}{\sqrt{\mathrm{T}}} \AA$
COMEDK 2013
Dual nature of radiation and Matter
142507
The de-Broglie wavelength of an electron, $\alpha$ particle and a proton all having the same kinetic energy is respectively given as $\lambda_{\mathrm{e}}, \lambda_{\alpha}$ and $\lambda_{\mathrm{p}}$. Then which of the following is not true?
142505
If the kinetic energy of free electron is made double, the new de-Broglie wave length will be times that of initial wave length.
1 $\sqrt{2}$
2 $\frac{1}{\sqrt{2}}$
3 2
4 $\frac{1}{2}$
Explanation:
B We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Kinetic energy of an moving electron, $\mathrm{K} =\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}^{2} =\frac{2 \mathrm{~K}}{\mathrm{~m}}$ Put the above value of $v$ in equation (i) we get, $\lambda=\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Case first:- $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}}}$ For second case:- $\mathrm{K}_{2}=2 \mathrm{~K}_{1}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{2}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times 2 \mathrm{~K}_{1}}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}} \times \sqrt{2}}$ Substituting $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{1}}}$ in the equation of $\lambda_{2}$ $\lambda_{2}=\frac{1}{\sqrt{2}} \lambda_{1}$ So, new wavelength is $\frac{1}{\sqrt{2}}$ times of the initial wavelength.
GUJCET 2014
Dual nature of radiation and Matter
142506
The de-Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is
1 $\frac{3.08}{\sqrt{\mathrm{T}}} \AA$
2 $\frac{0.308}{\sqrt{\mathrm{T}}} \AA$
3 $\frac{0.0308}{\sqrt{\mathrm{T}}} \AA$
4 $\frac{30.8}{\sqrt{\mathrm{T}}} \AA$
Explanation:
D Given that, $\mathrm{m}=1.67 \times 10^{-27} \mathrm{~kg}, \mathrm{~K}_{\mathrm{B}}=1.38 \times$ $10^{-23}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ We know that, de-Broglie wavelength of neutron at thermal equilibrium, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{\mathrm{B}} \mathrm{T}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times \mathrm{T}}}$ $\lambda=\frac{30.8 \times 10^{-10}}{\sqrt{\mathrm{T}} \mathrm{m}}$ $\lambda=\frac{30.8}{\sqrt{\mathrm{T}}} \AA$
COMEDK 2013
Dual nature of radiation and Matter
142507
The de-Broglie wavelength of an electron, $\alpha$ particle and a proton all having the same kinetic energy is respectively given as $\lambda_{\mathrm{e}}, \lambda_{\alpha}$ and $\lambda_{\mathrm{p}}$. Then which of the following is not true?
