142627
The distance between two successive atomic planes of a calcite crystal is \(0.3 \mathrm{~nm}\). The minimum angle for bragg scattering of \(0.3 \AA \mathrm{X}\) rays will be
1 \(1.43^{\circ}\)
2 \(1.56^{\circ}\)
3 \(2.86^{\circ}\)
4 \(30^{\circ}\)
Explanation:
C Given, Distance between two successive atomic planes \((\mathrm{d})=0.3 \mathrm{~nm}=0.3 \times 10^{-9} \mathrm{~m}\) \(\lambda=0.3 \AA=0.3 \times 10^{-10} \mathrm{~m}\) From Bragg's law \(\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta\) \(\mathrm{n}=1\), for minimum scattering \(0.3 \times 10^{-10}=2 \times 0.3 \times 10^{-9} \times \sin \theta\) \(\sin \theta=\frac{0.3 \times 10^{-10}}{2 \times 0.3 \times 10^{-9}}\) \(\sin \theta=\frac{1}{20}\) \(\theta=\sin ^{-1}\left(\frac{1}{20}\right)\) \(\theta=2.86^{\circ}\)
UP CPMT-2006
Dual nature of radiation and Matter
142630
Absorption of X-Rays is maximum in which of the following material short of same thickness ?
1 \(\mathrm{Cu}\)
2 \(\mathrm{Au}\)
3 \(\mathrm{Be}\)
4 \(\mathrm{Pb}\)
Explanation:
D We know Intensity (I) \(=\mathrm{I}_0 \mathrm{e}^{-\mu \mathrm{x}}\) Where, \(\mu=\) absorbtion co-efficient \(\mathrm{x}=\) Thickness of sheet \(\mathrm{I}_0=\) Initial Intensity And \(\mu \propto Z^2\) Where, \(\mathrm{Z}=\) atomic number Hence, Among four material \(\mathrm{Pb}\) has highest atomic number. So, \(\mathrm{Pb}\) has maximum absorption of \(\mathrm{X}\)-rays.
MP PMT-2009
Dual nature of radiation and Matter
142632
The minimum wavelength of \(\mathrm{X}\)-ray emitted by \(\mathrm{X}\)-ray tube is \(0.4125 \hat{\AA}\). The accelerating voltage is
142633
An isotropic point source emits light with wavelength \(500 \mathrm{~nm}\). The radiation power of the source is \(P=10 \mathrm{~W}\). Find the number of photons passing through unit area per second at a distance of \(3 \mathbf{~ m}\) from the source.
B Given, Wavelength \((\lambda)=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) Power \((\mathrm{P})=10 \mathrm{~W}\) We know, \(\mathrm{P}=\frac{\mathrm{n}_0 \mathrm{hc}}{\lambda}\) Number of photon per second \(\mathrm{n}_0=\frac{\mathrm{P} \lambda}{\mathrm{hc}}=\frac{10 \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}\) \(\mathrm{n}_0=2.5252 \times 10^{19} \mathrm{sec}\) Number of photon per unit area \(\mathrm{n}=\frac{\mathrm{n}_0}{4 \pi^2}=\frac{2.5252 \times 10^{19}}{4 \pi \times(3)^2}\) \(\mathrm{n}=2.23 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142627
The distance between two successive atomic planes of a calcite crystal is \(0.3 \mathrm{~nm}\). The minimum angle for bragg scattering of \(0.3 \AA \mathrm{X}\) rays will be
1 \(1.43^{\circ}\)
2 \(1.56^{\circ}\)
3 \(2.86^{\circ}\)
4 \(30^{\circ}\)
Explanation:
C Given, Distance between two successive atomic planes \((\mathrm{d})=0.3 \mathrm{~nm}=0.3 \times 10^{-9} \mathrm{~m}\) \(\lambda=0.3 \AA=0.3 \times 10^{-10} \mathrm{~m}\) From Bragg's law \(\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta\) \(\mathrm{n}=1\), for minimum scattering \(0.3 \times 10^{-10}=2 \times 0.3 \times 10^{-9} \times \sin \theta\) \(\sin \theta=\frac{0.3 \times 10^{-10}}{2 \times 0.3 \times 10^{-9}}\) \(\sin \theta=\frac{1}{20}\) \(\theta=\sin ^{-1}\left(\frac{1}{20}\right)\) \(\theta=2.86^{\circ}\)
UP CPMT-2006
Dual nature of radiation and Matter
142630
Absorption of X-Rays is maximum in which of the following material short of same thickness ?
1 \(\mathrm{Cu}\)
2 \(\mathrm{Au}\)
3 \(\mathrm{Be}\)
4 \(\mathrm{Pb}\)
Explanation:
D We know Intensity (I) \(=\mathrm{I}_0 \mathrm{e}^{-\mu \mathrm{x}}\) Where, \(\mu=\) absorbtion co-efficient \(\mathrm{x}=\) Thickness of sheet \(\mathrm{I}_0=\) Initial Intensity And \(\mu \propto Z^2\) Where, \(\mathrm{Z}=\) atomic number Hence, Among four material \(\mathrm{Pb}\) has highest atomic number. So, \(\mathrm{Pb}\) has maximum absorption of \(\mathrm{X}\)-rays.
MP PMT-2009
Dual nature of radiation and Matter
142632
The minimum wavelength of \(\mathrm{X}\)-ray emitted by \(\mathrm{X}\)-ray tube is \(0.4125 \hat{\AA}\). The accelerating voltage is
142633
An isotropic point source emits light with wavelength \(500 \mathrm{~nm}\). The radiation power of the source is \(P=10 \mathrm{~W}\). Find the number of photons passing through unit area per second at a distance of \(3 \mathbf{~ m}\) from the source.
B Given, Wavelength \((\lambda)=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) Power \((\mathrm{P})=10 \mathrm{~W}\) We know, \(\mathrm{P}=\frac{\mathrm{n}_0 \mathrm{hc}}{\lambda}\) Number of photon per second \(\mathrm{n}_0=\frac{\mathrm{P} \lambda}{\mathrm{hc}}=\frac{10 \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}\) \(\mathrm{n}_0=2.5252 \times 10^{19} \mathrm{sec}\) Number of photon per unit area \(\mathrm{n}=\frac{\mathrm{n}_0}{4 \pi^2}=\frac{2.5252 \times 10^{19}}{4 \pi \times(3)^2}\) \(\mathrm{n}=2.23 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}\)
142627
The distance between two successive atomic planes of a calcite crystal is \(0.3 \mathrm{~nm}\). The minimum angle for bragg scattering of \(0.3 \AA \mathrm{X}\) rays will be
1 \(1.43^{\circ}\)
2 \(1.56^{\circ}\)
3 \(2.86^{\circ}\)
4 \(30^{\circ}\)
Explanation:
C Given, Distance between two successive atomic planes \((\mathrm{d})=0.3 \mathrm{~nm}=0.3 \times 10^{-9} \mathrm{~m}\) \(\lambda=0.3 \AA=0.3 \times 10^{-10} \mathrm{~m}\) From Bragg's law \(\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta\) \(\mathrm{n}=1\), for minimum scattering \(0.3 \times 10^{-10}=2 \times 0.3 \times 10^{-9} \times \sin \theta\) \(\sin \theta=\frac{0.3 \times 10^{-10}}{2 \times 0.3 \times 10^{-9}}\) \(\sin \theta=\frac{1}{20}\) \(\theta=\sin ^{-1}\left(\frac{1}{20}\right)\) \(\theta=2.86^{\circ}\)
UP CPMT-2006
Dual nature of radiation and Matter
142630
Absorption of X-Rays is maximum in which of the following material short of same thickness ?
1 \(\mathrm{Cu}\)
2 \(\mathrm{Au}\)
3 \(\mathrm{Be}\)
4 \(\mathrm{Pb}\)
Explanation:
D We know Intensity (I) \(=\mathrm{I}_0 \mathrm{e}^{-\mu \mathrm{x}}\) Where, \(\mu=\) absorbtion co-efficient \(\mathrm{x}=\) Thickness of sheet \(\mathrm{I}_0=\) Initial Intensity And \(\mu \propto Z^2\) Where, \(\mathrm{Z}=\) atomic number Hence, Among four material \(\mathrm{Pb}\) has highest atomic number. So, \(\mathrm{Pb}\) has maximum absorption of \(\mathrm{X}\)-rays.
MP PMT-2009
Dual nature of radiation and Matter
142632
The minimum wavelength of \(\mathrm{X}\)-ray emitted by \(\mathrm{X}\)-ray tube is \(0.4125 \hat{\AA}\). The accelerating voltage is
142633
An isotropic point source emits light with wavelength \(500 \mathrm{~nm}\). The radiation power of the source is \(P=10 \mathrm{~W}\). Find the number of photons passing through unit area per second at a distance of \(3 \mathbf{~ m}\) from the source.
B Given, Wavelength \((\lambda)=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) Power \((\mathrm{P})=10 \mathrm{~W}\) We know, \(\mathrm{P}=\frac{\mathrm{n}_0 \mathrm{hc}}{\lambda}\) Number of photon per second \(\mathrm{n}_0=\frac{\mathrm{P} \lambda}{\mathrm{hc}}=\frac{10 \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}\) \(\mathrm{n}_0=2.5252 \times 10^{19} \mathrm{sec}\) Number of photon per unit area \(\mathrm{n}=\frac{\mathrm{n}_0}{4 \pi^2}=\frac{2.5252 \times 10^{19}}{4 \pi \times(3)^2}\) \(\mathrm{n}=2.23 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}\)
142627
The distance between two successive atomic planes of a calcite crystal is \(0.3 \mathrm{~nm}\). The minimum angle for bragg scattering of \(0.3 \AA \mathrm{X}\) rays will be
1 \(1.43^{\circ}\)
2 \(1.56^{\circ}\)
3 \(2.86^{\circ}\)
4 \(30^{\circ}\)
Explanation:
C Given, Distance between two successive atomic planes \((\mathrm{d})=0.3 \mathrm{~nm}=0.3 \times 10^{-9} \mathrm{~m}\) \(\lambda=0.3 \AA=0.3 \times 10^{-10} \mathrm{~m}\) From Bragg's law \(\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta\) \(\mathrm{n}=1\), for minimum scattering \(0.3 \times 10^{-10}=2 \times 0.3 \times 10^{-9} \times \sin \theta\) \(\sin \theta=\frac{0.3 \times 10^{-10}}{2 \times 0.3 \times 10^{-9}}\) \(\sin \theta=\frac{1}{20}\) \(\theta=\sin ^{-1}\left(\frac{1}{20}\right)\) \(\theta=2.86^{\circ}\)
UP CPMT-2006
Dual nature of radiation and Matter
142630
Absorption of X-Rays is maximum in which of the following material short of same thickness ?
1 \(\mathrm{Cu}\)
2 \(\mathrm{Au}\)
3 \(\mathrm{Be}\)
4 \(\mathrm{Pb}\)
Explanation:
D We know Intensity (I) \(=\mathrm{I}_0 \mathrm{e}^{-\mu \mathrm{x}}\) Where, \(\mu=\) absorbtion co-efficient \(\mathrm{x}=\) Thickness of sheet \(\mathrm{I}_0=\) Initial Intensity And \(\mu \propto Z^2\) Where, \(\mathrm{Z}=\) atomic number Hence, Among four material \(\mathrm{Pb}\) has highest atomic number. So, \(\mathrm{Pb}\) has maximum absorption of \(\mathrm{X}\)-rays.
MP PMT-2009
Dual nature of radiation and Matter
142632
The minimum wavelength of \(\mathrm{X}\)-ray emitted by \(\mathrm{X}\)-ray tube is \(0.4125 \hat{\AA}\). The accelerating voltage is
142633
An isotropic point source emits light with wavelength \(500 \mathrm{~nm}\). The radiation power of the source is \(P=10 \mathrm{~W}\). Find the number of photons passing through unit area per second at a distance of \(3 \mathbf{~ m}\) from the source.
B Given, Wavelength \((\lambda)=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) Power \((\mathrm{P})=10 \mathrm{~W}\) We know, \(\mathrm{P}=\frac{\mathrm{n}_0 \mathrm{hc}}{\lambda}\) Number of photon per second \(\mathrm{n}_0=\frac{\mathrm{P} \lambda}{\mathrm{hc}}=\frac{10 \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}\) \(\mathrm{n}_0=2.5252 \times 10^{19} \mathrm{sec}\) Number of photon per unit area \(\mathrm{n}=\frac{\mathrm{n}_0}{4 \pi^2}=\frac{2.5252 \times 10^{19}}{4 \pi \times(3)^2}\) \(\mathrm{n}=2.23 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}\)