NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142396
The momentum of a proton, a neutron and an electron are in the ratio $3: 2: 1$, then their respective de Broglie wavelengths are in the ratio
1 $1: 1: 1$
2 $2: 3: 6$
3 $1: 2: 3$
4 $6: 3: 2$
5 $4: 2: 1$
Explanation:
B Given that, The ratio of momentum of a proton, neutron and electron are- We know, $p_{p}: p_{n}: p_{e}=3: 2: 1$ $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda \propto \frac{1}{\mathrm{p}}$ $\therefore \quad \lambda_{\mathrm{p}}: \lambda_{\mathrm{b}}: \lambda_{\mathrm{e}} =\frac{1}{3}: \frac{1}{2}: 1$ $=\frac{1}{3} \times 6: \frac{1}{2} \times 6: 1 \times 6$ $=2: 3: 6$
Kerala CEE 04.07.2022
Dual nature of radiation and Matter
142397
A particle of mass $1 \times 10^{-30} \mathrm{~kg}$ and electric change $1.6 \times 10^{-19} \mathrm{C}$ has de-Broglie wavelength $660 \mathrm{~nm}$. Then kinetic energy of this particle is-
142398
The mass of a particle is $\mathbf{1 6}$ times heavier than another particle and they have the same kinetic energy. The de-Broglie wavelength of the lighter particle is
1 Same as the heavier particle's wavelength
2 2 times the heavier particle's wavelength
3 4 times the heavier particle's wavelength
4 1/4 times the heavier particle's wavelength
Explanation:
C Given, $\mathrm{K}_{1} =\mathrm{K}_{2}=\mathrm{K}$ $\mathrm{m}_{1} =\mathrm{m}, \mathrm{m}_{2}=16 \mathrm{~m}$ We know, de - Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{~K}_{1}}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{~K}_{2}}}$ Equation (i) divided by equation (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{~K}_{1}}}}{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{~K}_{2}}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{m}_{2} \mathrm{~K}_{2}}{\mathrm{~m}_{1} \mathrm{~K}_{1}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{16 \mathrm{mK}}{\mathrm{mK}}}=\sqrt{16}$ $\frac{\lambda_{1}}{\lambda_{2}}=4$ $\lambda_{1}=4 \lambda_{2}$
Shift-II
Dual nature of radiation and Matter
142399
An electron in the hydrogen atom excites from $2^{\text {nd }}$ orbit to $4^{\text {th }}$ orbit then the change in angular momentum of the electron is \(\left(\right.\) Planck's constant \(\left.h=6.64 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\)
1 $2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
2 $1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
3 $0.57 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
4 $4.22 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
Explanation:
A Change in the angular momentum $\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}$ We know that, $\mathrm{L}_{1}=\frac{\mathrm{n}_{1} \mathrm{~h}}{2 \pi}$ $\mathrm{L}_{2}=\frac{\mathrm{n}_{2} \mathrm{~h}}{2 \pi}$ Putting the value of $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ from equation (ii) and (iii) in equation (i) $\Delta \mathrm{L}=\frac{\mathrm{n}_{2} \mathrm{~h}}{2 \pi}-\frac{\mathrm{n}_{1} \mathrm{~h}}{2 \pi}$ $\Delta \mathrm{L} =\frac{\mathrm{h}}{2 \pi}\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)$ $\Delta \mathrm{L} =\frac{6.64 \times 10^{-34}}{2 \times 3.14}(4-2)$ $\Delta \mathrm{L} =2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
142396
The momentum of a proton, a neutron and an electron are in the ratio $3: 2: 1$, then their respective de Broglie wavelengths are in the ratio
1 $1: 1: 1$
2 $2: 3: 6$
3 $1: 2: 3$
4 $6: 3: 2$
5 $4: 2: 1$
Explanation:
B Given that, The ratio of momentum of a proton, neutron and electron are- We know, $p_{p}: p_{n}: p_{e}=3: 2: 1$ $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda \propto \frac{1}{\mathrm{p}}$ $\therefore \quad \lambda_{\mathrm{p}}: \lambda_{\mathrm{b}}: \lambda_{\mathrm{e}} =\frac{1}{3}: \frac{1}{2}: 1$ $=\frac{1}{3} \times 6: \frac{1}{2} \times 6: 1 \times 6$ $=2: 3: 6$
Kerala CEE 04.07.2022
Dual nature of radiation and Matter
142397
A particle of mass $1 \times 10^{-30} \mathrm{~kg}$ and electric change $1.6 \times 10^{-19} \mathrm{C}$ has de-Broglie wavelength $660 \mathrm{~nm}$. Then kinetic energy of this particle is-
142398
The mass of a particle is $\mathbf{1 6}$ times heavier than another particle and they have the same kinetic energy. The de-Broglie wavelength of the lighter particle is
1 Same as the heavier particle's wavelength
2 2 times the heavier particle's wavelength
3 4 times the heavier particle's wavelength
4 1/4 times the heavier particle's wavelength
Explanation:
C Given, $\mathrm{K}_{1} =\mathrm{K}_{2}=\mathrm{K}$ $\mathrm{m}_{1} =\mathrm{m}, \mathrm{m}_{2}=16 \mathrm{~m}$ We know, de - Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{~K}_{1}}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{~K}_{2}}}$ Equation (i) divided by equation (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{~K}_{1}}}}{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{~K}_{2}}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{m}_{2} \mathrm{~K}_{2}}{\mathrm{~m}_{1} \mathrm{~K}_{1}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{16 \mathrm{mK}}{\mathrm{mK}}}=\sqrt{16}$ $\frac{\lambda_{1}}{\lambda_{2}}=4$ $\lambda_{1}=4 \lambda_{2}$
Shift-II
Dual nature of radiation and Matter
142399
An electron in the hydrogen atom excites from $2^{\text {nd }}$ orbit to $4^{\text {th }}$ orbit then the change in angular momentum of the electron is \(\left(\right.\) Planck's constant \(\left.h=6.64 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\)
1 $2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
2 $1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
3 $0.57 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
4 $4.22 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
Explanation:
A Change in the angular momentum $\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}$ We know that, $\mathrm{L}_{1}=\frac{\mathrm{n}_{1} \mathrm{~h}}{2 \pi}$ $\mathrm{L}_{2}=\frac{\mathrm{n}_{2} \mathrm{~h}}{2 \pi}$ Putting the value of $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ from equation (ii) and (iii) in equation (i) $\Delta \mathrm{L}=\frac{\mathrm{n}_{2} \mathrm{~h}}{2 \pi}-\frac{\mathrm{n}_{1} \mathrm{~h}}{2 \pi}$ $\Delta \mathrm{L} =\frac{\mathrm{h}}{2 \pi}\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)$ $\Delta \mathrm{L} =\frac{6.64 \times 10^{-34}}{2 \times 3.14}(4-2)$ $\Delta \mathrm{L} =2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
142396
The momentum of a proton, a neutron and an electron are in the ratio $3: 2: 1$, then their respective de Broglie wavelengths are in the ratio
1 $1: 1: 1$
2 $2: 3: 6$
3 $1: 2: 3$
4 $6: 3: 2$
5 $4: 2: 1$
Explanation:
B Given that, The ratio of momentum of a proton, neutron and electron are- We know, $p_{p}: p_{n}: p_{e}=3: 2: 1$ $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda \propto \frac{1}{\mathrm{p}}$ $\therefore \quad \lambda_{\mathrm{p}}: \lambda_{\mathrm{b}}: \lambda_{\mathrm{e}} =\frac{1}{3}: \frac{1}{2}: 1$ $=\frac{1}{3} \times 6: \frac{1}{2} \times 6: 1 \times 6$ $=2: 3: 6$
Kerala CEE 04.07.2022
Dual nature of radiation and Matter
142397
A particle of mass $1 \times 10^{-30} \mathrm{~kg}$ and electric change $1.6 \times 10^{-19} \mathrm{C}$ has de-Broglie wavelength $660 \mathrm{~nm}$. Then kinetic energy of this particle is-
142398
The mass of a particle is $\mathbf{1 6}$ times heavier than another particle and they have the same kinetic energy. The de-Broglie wavelength of the lighter particle is
1 Same as the heavier particle's wavelength
2 2 times the heavier particle's wavelength
3 4 times the heavier particle's wavelength
4 1/4 times the heavier particle's wavelength
Explanation:
C Given, $\mathrm{K}_{1} =\mathrm{K}_{2}=\mathrm{K}$ $\mathrm{m}_{1} =\mathrm{m}, \mathrm{m}_{2}=16 \mathrm{~m}$ We know, de - Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{~K}_{1}}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{~K}_{2}}}$ Equation (i) divided by equation (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{~K}_{1}}}}{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{~K}_{2}}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{m}_{2} \mathrm{~K}_{2}}{\mathrm{~m}_{1} \mathrm{~K}_{1}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{16 \mathrm{mK}}{\mathrm{mK}}}=\sqrt{16}$ $\frac{\lambda_{1}}{\lambda_{2}}=4$ $\lambda_{1}=4 \lambda_{2}$
Shift-II
Dual nature of radiation and Matter
142399
An electron in the hydrogen atom excites from $2^{\text {nd }}$ orbit to $4^{\text {th }}$ orbit then the change in angular momentum of the electron is \(\left(\right.\) Planck's constant \(\left.h=6.64 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\)
1 $2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
2 $1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
3 $0.57 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
4 $4.22 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
Explanation:
A Change in the angular momentum $\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}$ We know that, $\mathrm{L}_{1}=\frac{\mathrm{n}_{1} \mathrm{~h}}{2 \pi}$ $\mathrm{L}_{2}=\frac{\mathrm{n}_{2} \mathrm{~h}}{2 \pi}$ Putting the value of $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ from equation (ii) and (iii) in equation (i) $\Delta \mathrm{L}=\frac{\mathrm{n}_{2} \mathrm{~h}}{2 \pi}-\frac{\mathrm{n}_{1} \mathrm{~h}}{2 \pi}$ $\Delta \mathrm{L} =\frac{\mathrm{h}}{2 \pi}\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)$ $\Delta \mathrm{L} =\frac{6.64 \times 10^{-34}}{2 \times 3.14}(4-2)$ $\Delta \mathrm{L} =2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Dual nature of radiation and Matter
142396
The momentum of a proton, a neutron and an electron are in the ratio $3: 2: 1$, then their respective de Broglie wavelengths are in the ratio
1 $1: 1: 1$
2 $2: 3: 6$
3 $1: 2: 3$
4 $6: 3: 2$
5 $4: 2: 1$
Explanation:
B Given that, The ratio of momentum of a proton, neutron and electron are- We know, $p_{p}: p_{n}: p_{e}=3: 2: 1$ $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda \propto \frac{1}{\mathrm{p}}$ $\therefore \quad \lambda_{\mathrm{p}}: \lambda_{\mathrm{b}}: \lambda_{\mathrm{e}} =\frac{1}{3}: \frac{1}{2}: 1$ $=\frac{1}{3} \times 6: \frac{1}{2} \times 6: 1 \times 6$ $=2: 3: 6$
Kerala CEE 04.07.2022
Dual nature of radiation and Matter
142397
A particle of mass $1 \times 10^{-30} \mathrm{~kg}$ and electric change $1.6 \times 10^{-19} \mathrm{C}$ has de-Broglie wavelength $660 \mathrm{~nm}$. Then kinetic energy of this particle is-
142398
The mass of a particle is $\mathbf{1 6}$ times heavier than another particle and they have the same kinetic energy. The de-Broglie wavelength of the lighter particle is
1 Same as the heavier particle's wavelength
2 2 times the heavier particle's wavelength
3 4 times the heavier particle's wavelength
4 1/4 times the heavier particle's wavelength
Explanation:
C Given, $\mathrm{K}_{1} =\mathrm{K}_{2}=\mathrm{K}$ $\mathrm{m}_{1} =\mathrm{m}, \mathrm{m}_{2}=16 \mathrm{~m}$ We know, de - Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{~K}_{1}}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{~K}_{2}}}$ Equation (i) divided by equation (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{~K}_{1}}}}{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{~K}_{2}}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{m}_{2} \mathrm{~K}_{2}}{\mathrm{~m}_{1} \mathrm{~K}_{1}}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{16 \mathrm{mK}}{\mathrm{mK}}}=\sqrt{16}$ $\frac{\lambda_{1}}{\lambda_{2}}=4$ $\lambda_{1}=4 \lambda_{2}$
Shift-II
Dual nature of radiation and Matter
142399
An electron in the hydrogen atom excites from $2^{\text {nd }}$ orbit to $4^{\text {th }}$ orbit then the change in angular momentum of the electron is \(\left(\right.\) Planck's constant \(\left.h=6.64 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\)
1 $2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
2 $1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
3 $0.57 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
4 $4.22 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
Explanation:
A Change in the angular momentum $\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}$ We know that, $\mathrm{L}_{1}=\frac{\mathrm{n}_{1} \mathrm{~h}}{2 \pi}$ $\mathrm{L}_{2}=\frac{\mathrm{n}_{2} \mathrm{~h}}{2 \pi}$ Putting the value of $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ from equation (ii) and (iii) in equation (i) $\Delta \mathrm{L}=\frac{\mathrm{n}_{2} \mathrm{~h}}{2 \pi}-\frac{\mathrm{n}_{1} \mathrm{~h}}{2 \pi}$ $\Delta \mathrm{L} =\frac{\mathrm{h}}{2 \pi}\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)$ $\Delta \mathrm{L} =\frac{6.64 \times 10^{-34}}{2 \times 3.14}(4-2)$ $\Delta \mathrm{L} =2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$