142468
In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage $V$ then de-Broglie wavelength of that electron will be m.
1 $\frac{2 \mathrm{Vem}}{\sqrt{\mathrm{h}}}$
2 $\frac{\sqrt{\mathrm{h}}}{2 \mathrm{Vem}}$
3 $\frac{\sqrt{2 \mathrm{Vem}}}{\mathrm{h}}$
4 $\frac{\mathrm{h}}{\sqrt{2 \mathrm{Vem}}}$
Explanation:
D We know that, Kinetic energy of electron due to applied voltage, Since, $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p} =\mathrm{mv}$ $\therefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\text { So, } \quad \lambda =\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{eVm}}}$
GUJCET 2017
Dual nature of radiation and Matter
142469
The de-Broglie wavelength of a proton (mass $=$ $1.6 \times 10^{-27} \mathrm{~kg}$ ) accelerated through a potential difference of $1 \mathrm{kV}$ is
1 $600 \AA$
2 $0.9 \times 10^{-12} \mathrm{~m}$
3 $7 \AA$
4 $0.9 \mathrm{~mm}$.
Explanation:
B Given that, $\mathrm{V}=1 \mathrm{kV}=1 \times 10^{3} \mathrm{~V}, \mathrm{~m}=1.6 \times$ $10^{-27} \mathrm{~kg}$ Since, the proton is accelerated and moving, it kinetic energy, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ $\mathrm{p}=\sqrt{2 \mathrm{mKE}}$ We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Putting the value of $p$ from equation (i) in equation (ii) $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-27}\left(1.6 \times 10^{-19} \times 1 \times 10^{3}\right)}}$ $\lambda=0.9 \times 10^{-12} \mathrm{~m}$
BITSAT-2017
Dual nature of radiation and Matter
142470
The de-Broglie wavelength of a neutron at $27^{\circ} \mathrm{C}$ is $\lambda$. What will be its wavelength at $927^{\circ} \mathrm{C}$ ?
1 $\lambda / 2$
2 $\lambda / 3$
3 $\lambda / 4$
4 $\lambda / 9$
Explanation:
A Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}, \mathrm{T}_{2}=927^{\circ} \mathrm{C}$ The wavelength is inversely proportional to square root of temperature. $\lambda_{\text {neutron }} \propto \frac{1}{\sqrt{\mathrm{T}}}$ $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}$ $\text { Let, } \lambda_{1}=\lambda$ Putting these value, we get - $\frac{\lambda_{1}}{\lambda_{2}} =\sqrt{\frac{(273+927)}{(273+27)}}$ $\frac{\lambda}{\lambda_{2}} =\sqrt{\frac{1200}{300}}=\sqrt{4}=2$ $\therefore \quad \lambda_{2} =\frac{\lambda}{2}$
Manipal UGET-2014
Dual nature of radiation and Matter
142471
de - Broglie wavelength of a body of mass $1 \mathrm{~kg}$ moving with velocity of $2000 \mathrm{~m} / \mathrm{s}$ is
1 $3.32 \times 10^{-27} \AA$
2 $1.5 \times 10^{7} \AA$
3 $0.55 \times 10^{-22} \AA$
4 None of these
Explanation:
A Given that, mass $=1 \mathrm{~kg}, \mathrm{v}=2000 \mathrm{~m} / \mathrm{s}$ We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.6 \times 10^{-34}}{1 \times 2000}$ $\lambda=3.3 \times 10^{-37} \mathrm{~m}$ $\lambda=3.3 \times 10^{-27} \AA$
142468
In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage $V$ then de-Broglie wavelength of that electron will be m.
1 $\frac{2 \mathrm{Vem}}{\sqrt{\mathrm{h}}}$
2 $\frac{\sqrt{\mathrm{h}}}{2 \mathrm{Vem}}$
3 $\frac{\sqrt{2 \mathrm{Vem}}}{\mathrm{h}}$
4 $\frac{\mathrm{h}}{\sqrt{2 \mathrm{Vem}}}$
Explanation:
D We know that, Kinetic energy of electron due to applied voltage, Since, $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p} =\mathrm{mv}$ $\therefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\text { So, } \quad \lambda =\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{eVm}}}$
GUJCET 2017
Dual nature of radiation and Matter
142469
The de-Broglie wavelength of a proton (mass $=$ $1.6 \times 10^{-27} \mathrm{~kg}$ ) accelerated through a potential difference of $1 \mathrm{kV}$ is
1 $600 \AA$
2 $0.9 \times 10^{-12} \mathrm{~m}$
3 $7 \AA$
4 $0.9 \mathrm{~mm}$.
Explanation:
B Given that, $\mathrm{V}=1 \mathrm{kV}=1 \times 10^{3} \mathrm{~V}, \mathrm{~m}=1.6 \times$ $10^{-27} \mathrm{~kg}$ Since, the proton is accelerated and moving, it kinetic energy, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ $\mathrm{p}=\sqrt{2 \mathrm{mKE}}$ We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Putting the value of $p$ from equation (i) in equation (ii) $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-27}\left(1.6 \times 10^{-19} \times 1 \times 10^{3}\right)}}$ $\lambda=0.9 \times 10^{-12} \mathrm{~m}$
BITSAT-2017
Dual nature of radiation and Matter
142470
The de-Broglie wavelength of a neutron at $27^{\circ} \mathrm{C}$ is $\lambda$. What will be its wavelength at $927^{\circ} \mathrm{C}$ ?
1 $\lambda / 2$
2 $\lambda / 3$
3 $\lambda / 4$
4 $\lambda / 9$
Explanation:
A Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}, \mathrm{T}_{2}=927^{\circ} \mathrm{C}$ The wavelength is inversely proportional to square root of temperature. $\lambda_{\text {neutron }} \propto \frac{1}{\sqrt{\mathrm{T}}}$ $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}$ $\text { Let, } \lambda_{1}=\lambda$ Putting these value, we get - $\frac{\lambda_{1}}{\lambda_{2}} =\sqrt{\frac{(273+927)}{(273+27)}}$ $\frac{\lambda}{\lambda_{2}} =\sqrt{\frac{1200}{300}}=\sqrt{4}=2$ $\therefore \quad \lambda_{2} =\frac{\lambda}{2}$
Manipal UGET-2014
Dual nature of radiation and Matter
142471
de - Broglie wavelength of a body of mass $1 \mathrm{~kg}$ moving with velocity of $2000 \mathrm{~m} / \mathrm{s}$ is
1 $3.32 \times 10^{-27} \AA$
2 $1.5 \times 10^{7} \AA$
3 $0.55 \times 10^{-22} \AA$
4 None of these
Explanation:
A Given that, mass $=1 \mathrm{~kg}, \mathrm{v}=2000 \mathrm{~m} / \mathrm{s}$ We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.6 \times 10^{-34}}{1 \times 2000}$ $\lambda=3.3 \times 10^{-37} \mathrm{~m}$ $\lambda=3.3 \times 10^{-27} \AA$
142468
In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage $V$ then de-Broglie wavelength of that electron will be m.
1 $\frac{2 \mathrm{Vem}}{\sqrt{\mathrm{h}}}$
2 $\frac{\sqrt{\mathrm{h}}}{2 \mathrm{Vem}}$
3 $\frac{\sqrt{2 \mathrm{Vem}}}{\mathrm{h}}$
4 $\frac{\mathrm{h}}{\sqrt{2 \mathrm{Vem}}}$
Explanation:
D We know that, Kinetic energy of electron due to applied voltage, Since, $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p} =\mathrm{mv}$ $\therefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\text { So, } \quad \lambda =\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{eVm}}}$
GUJCET 2017
Dual nature of radiation and Matter
142469
The de-Broglie wavelength of a proton (mass $=$ $1.6 \times 10^{-27} \mathrm{~kg}$ ) accelerated through a potential difference of $1 \mathrm{kV}$ is
1 $600 \AA$
2 $0.9 \times 10^{-12} \mathrm{~m}$
3 $7 \AA$
4 $0.9 \mathrm{~mm}$.
Explanation:
B Given that, $\mathrm{V}=1 \mathrm{kV}=1 \times 10^{3} \mathrm{~V}, \mathrm{~m}=1.6 \times$ $10^{-27} \mathrm{~kg}$ Since, the proton is accelerated and moving, it kinetic energy, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ $\mathrm{p}=\sqrt{2 \mathrm{mKE}}$ We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Putting the value of $p$ from equation (i) in equation (ii) $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-27}\left(1.6 \times 10^{-19} \times 1 \times 10^{3}\right)}}$ $\lambda=0.9 \times 10^{-12} \mathrm{~m}$
BITSAT-2017
Dual nature of radiation and Matter
142470
The de-Broglie wavelength of a neutron at $27^{\circ} \mathrm{C}$ is $\lambda$. What will be its wavelength at $927^{\circ} \mathrm{C}$ ?
1 $\lambda / 2$
2 $\lambda / 3$
3 $\lambda / 4$
4 $\lambda / 9$
Explanation:
A Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}, \mathrm{T}_{2}=927^{\circ} \mathrm{C}$ The wavelength is inversely proportional to square root of temperature. $\lambda_{\text {neutron }} \propto \frac{1}{\sqrt{\mathrm{T}}}$ $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}$ $\text { Let, } \lambda_{1}=\lambda$ Putting these value, we get - $\frac{\lambda_{1}}{\lambda_{2}} =\sqrt{\frac{(273+927)}{(273+27)}}$ $\frac{\lambda}{\lambda_{2}} =\sqrt{\frac{1200}{300}}=\sqrt{4}=2$ $\therefore \quad \lambda_{2} =\frac{\lambda}{2}$
Manipal UGET-2014
Dual nature of radiation and Matter
142471
de - Broglie wavelength of a body of mass $1 \mathrm{~kg}$ moving with velocity of $2000 \mathrm{~m} / \mathrm{s}$ is
1 $3.32 \times 10^{-27} \AA$
2 $1.5 \times 10^{7} \AA$
3 $0.55 \times 10^{-22} \AA$
4 None of these
Explanation:
A Given that, mass $=1 \mathrm{~kg}, \mathrm{v}=2000 \mathrm{~m} / \mathrm{s}$ We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.6 \times 10^{-34}}{1 \times 2000}$ $\lambda=3.3 \times 10^{-37} \mathrm{~m}$ $\lambda=3.3 \times 10^{-27} \AA$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142468
In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage $V$ then de-Broglie wavelength of that electron will be m.
1 $\frac{2 \mathrm{Vem}}{\sqrt{\mathrm{h}}}$
2 $\frac{\sqrt{\mathrm{h}}}{2 \mathrm{Vem}}$
3 $\frac{\sqrt{2 \mathrm{Vem}}}{\mathrm{h}}$
4 $\frac{\mathrm{h}}{\sqrt{2 \mathrm{Vem}}}$
Explanation:
D We know that, Kinetic energy of electron due to applied voltage, Since, $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p} =\mathrm{mv}$ $\therefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\text { So, } \quad \lambda =\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{eVm}}}$
GUJCET 2017
Dual nature of radiation and Matter
142469
The de-Broglie wavelength of a proton (mass $=$ $1.6 \times 10^{-27} \mathrm{~kg}$ ) accelerated through a potential difference of $1 \mathrm{kV}$ is
1 $600 \AA$
2 $0.9 \times 10^{-12} \mathrm{~m}$
3 $7 \AA$
4 $0.9 \mathrm{~mm}$.
Explanation:
B Given that, $\mathrm{V}=1 \mathrm{kV}=1 \times 10^{3} \mathrm{~V}, \mathrm{~m}=1.6 \times$ $10^{-27} \mathrm{~kg}$ Since, the proton is accelerated and moving, it kinetic energy, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ $\mathrm{p}=\sqrt{2 \mathrm{mKE}}$ We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Putting the value of $p$ from equation (i) in equation (ii) $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-27}\left(1.6 \times 10^{-19} \times 1 \times 10^{3}\right)}}$ $\lambda=0.9 \times 10^{-12} \mathrm{~m}$
BITSAT-2017
Dual nature of radiation and Matter
142470
The de-Broglie wavelength of a neutron at $27^{\circ} \mathrm{C}$ is $\lambda$. What will be its wavelength at $927^{\circ} \mathrm{C}$ ?
1 $\lambda / 2$
2 $\lambda / 3$
3 $\lambda / 4$
4 $\lambda / 9$
Explanation:
A Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}, \mathrm{T}_{2}=927^{\circ} \mathrm{C}$ The wavelength is inversely proportional to square root of temperature. $\lambda_{\text {neutron }} \propto \frac{1}{\sqrt{\mathrm{T}}}$ $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}$ $\text { Let, } \lambda_{1}=\lambda$ Putting these value, we get - $\frac{\lambda_{1}}{\lambda_{2}} =\sqrt{\frac{(273+927)}{(273+27)}}$ $\frac{\lambda}{\lambda_{2}} =\sqrt{\frac{1200}{300}}=\sqrt{4}=2$ $\therefore \quad \lambda_{2} =\frac{\lambda}{2}$
Manipal UGET-2014
Dual nature of radiation and Matter
142471
de - Broglie wavelength of a body of mass $1 \mathrm{~kg}$ moving with velocity of $2000 \mathrm{~m} / \mathrm{s}$ is
1 $3.32 \times 10^{-27} \AA$
2 $1.5 \times 10^{7} \AA$
3 $0.55 \times 10^{-22} \AA$
4 None of these
Explanation:
A Given that, mass $=1 \mathrm{~kg}, \mathrm{v}=2000 \mathrm{~m} / \mathrm{s}$ We know that, de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.6 \times 10^{-34}}{1 \times 2000}$ $\lambda=3.3 \times 10^{-37} \mathrm{~m}$ $\lambda=3.3 \times 10^{-27} \AA$
