NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142373
The momentum of electrons having a wavelength $2 \AA$ (Given $h=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{m}=$ $9.1 \times 10^{-35} \mathrm{~kg}$ ) is
B Given that, Wavelength $(\lambda)=2 \AA=2 \times 10^{-10} \mathrm{~m}$ Planck's constant $(h)=6.626 \times 10^{-34} \mathrm{Js}$ Let $\mathrm{p}$ be the momentum of the particle. We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{6.62 \times 10^{-34}}{2 \times 10^{-10}}$ $\mathrm{p}=3.31 \times 10^{-24} \mathrm{~kg}^{-\mathrm{ms}^{-1}}$
COMEDK 2015
Dual nature of radiation and Matter
142378
Newton postulated his corpuscular theory on the basis of :
1 Newton's rings
2 colours of thin films
3 rectilinear propagation of light
4 dispersion white light
Explanation:
C Newton postulated his corpuscular theory on the basic of following postulates are- (1) Newton proposed that a source of light emits many minutes particles, elastic, rigid and mass less particles called corpuscles. (2) The particles travel through transparent medium at very high speed in all direction along a straight line. (3) The corpuscles enter our eyes and produces the sensation of vision. (4) Due to different size of the corpuscles they produce different colours. (5) These light particles are replaced by a reflecting surface and attracted by transparent materials. Hence, it explain the rectilinear propagation of light.
Karnataka CET-2001
Dual nature of radiation and Matter
142379
The radius of a nucleus with atomic mass number 7 is 2 fermi. The radius of nucleus with atomic mass number 189 is
1 3 fermi
2 4 fermi
3 5 fermi
4 6 fermi
Explanation:
D Given that, $A_{1}=7, A_{2}=189, R_{1}=2$ fermi, $R_{2}$ $=$ ? The expression for nuclear radius is - Then, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{2}{\mathrm{R}_{2}}=\left(\frac{7}{189}\right)^{1 / 3} \Rightarrow \frac{2}{\mathrm{R}_{2}}=\left(\frac{1}{27}\right)^{1 / 3}$ $\frac{2}{\mathrm{R}_{2}}=\frac{1}{3}$ $\mathrm{R}=6 \text { Fermi }$ So,
J and K CET- 2007
Dual nature of radiation and Matter
142384
The X-rays spectrum coming from an X-ray tube
1 no minimum or maximum wavelengths
2 has all wavelength smaller than a certain maximum wavelength
3 has all wavelengths greater than a certain minimum wavelength
4 is monochromatic
Explanation:
C The X-rays spectrum coming from on X-ray tube has all wavelength greater than a certain minimum wavelength. X-rays are electromagnetic waves with wavelength range of $0.1 \AA-100 \AA$. $\mathrm{X}$-ray coming from the tube contains more than one wavelength therefore it is not monochromatic. Let the potential difference is V. Maximum energy of photon, is given by $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}=\mathrm{h} v_{\text {max }}=\frac{\mathrm{hc}}{\lambda_{\text {max }}}$ $\lambda=\frac{\mathrm{hc}}{\mathrm{eV}}$ X-ray beam has energy less than or equal to $E_{\max }$. Hence, wavelength greater than a certain minimum wavelength.
142373
The momentum of electrons having a wavelength $2 \AA$ (Given $h=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{m}=$ $9.1 \times 10^{-35} \mathrm{~kg}$ ) is
B Given that, Wavelength $(\lambda)=2 \AA=2 \times 10^{-10} \mathrm{~m}$ Planck's constant $(h)=6.626 \times 10^{-34} \mathrm{Js}$ Let $\mathrm{p}$ be the momentum of the particle. We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{6.62 \times 10^{-34}}{2 \times 10^{-10}}$ $\mathrm{p}=3.31 \times 10^{-24} \mathrm{~kg}^{-\mathrm{ms}^{-1}}$
COMEDK 2015
Dual nature of radiation and Matter
142378
Newton postulated his corpuscular theory on the basis of :
1 Newton's rings
2 colours of thin films
3 rectilinear propagation of light
4 dispersion white light
Explanation:
C Newton postulated his corpuscular theory on the basic of following postulates are- (1) Newton proposed that a source of light emits many minutes particles, elastic, rigid and mass less particles called corpuscles. (2) The particles travel through transparent medium at very high speed in all direction along a straight line. (3) The corpuscles enter our eyes and produces the sensation of vision. (4) Due to different size of the corpuscles they produce different colours. (5) These light particles are replaced by a reflecting surface and attracted by transparent materials. Hence, it explain the rectilinear propagation of light.
Karnataka CET-2001
Dual nature of radiation and Matter
142379
The radius of a nucleus with atomic mass number 7 is 2 fermi. The radius of nucleus with atomic mass number 189 is
1 3 fermi
2 4 fermi
3 5 fermi
4 6 fermi
Explanation:
D Given that, $A_{1}=7, A_{2}=189, R_{1}=2$ fermi, $R_{2}$ $=$ ? The expression for nuclear radius is - Then, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{2}{\mathrm{R}_{2}}=\left(\frac{7}{189}\right)^{1 / 3} \Rightarrow \frac{2}{\mathrm{R}_{2}}=\left(\frac{1}{27}\right)^{1 / 3}$ $\frac{2}{\mathrm{R}_{2}}=\frac{1}{3}$ $\mathrm{R}=6 \text { Fermi }$ So,
J and K CET- 2007
Dual nature of radiation and Matter
142384
The X-rays spectrum coming from an X-ray tube
1 no minimum or maximum wavelengths
2 has all wavelength smaller than a certain maximum wavelength
3 has all wavelengths greater than a certain minimum wavelength
4 is monochromatic
Explanation:
C The X-rays spectrum coming from on X-ray tube has all wavelength greater than a certain minimum wavelength. X-rays are electromagnetic waves with wavelength range of $0.1 \AA-100 \AA$. $\mathrm{X}$-ray coming from the tube contains more than one wavelength therefore it is not monochromatic. Let the potential difference is V. Maximum energy of photon, is given by $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}=\mathrm{h} v_{\text {max }}=\frac{\mathrm{hc}}{\lambda_{\text {max }}}$ $\lambda=\frac{\mathrm{hc}}{\mathrm{eV}}$ X-ray beam has energy less than or equal to $E_{\max }$. Hence, wavelength greater than a certain minimum wavelength.
142373
The momentum of electrons having a wavelength $2 \AA$ (Given $h=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{m}=$ $9.1 \times 10^{-35} \mathrm{~kg}$ ) is
B Given that, Wavelength $(\lambda)=2 \AA=2 \times 10^{-10} \mathrm{~m}$ Planck's constant $(h)=6.626 \times 10^{-34} \mathrm{Js}$ Let $\mathrm{p}$ be the momentum of the particle. We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{6.62 \times 10^{-34}}{2 \times 10^{-10}}$ $\mathrm{p}=3.31 \times 10^{-24} \mathrm{~kg}^{-\mathrm{ms}^{-1}}$
COMEDK 2015
Dual nature of radiation and Matter
142378
Newton postulated his corpuscular theory on the basis of :
1 Newton's rings
2 colours of thin films
3 rectilinear propagation of light
4 dispersion white light
Explanation:
C Newton postulated his corpuscular theory on the basic of following postulates are- (1) Newton proposed that a source of light emits many minutes particles, elastic, rigid and mass less particles called corpuscles. (2) The particles travel through transparent medium at very high speed in all direction along a straight line. (3) The corpuscles enter our eyes and produces the sensation of vision. (4) Due to different size of the corpuscles they produce different colours. (5) These light particles are replaced by a reflecting surface and attracted by transparent materials. Hence, it explain the rectilinear propagation of light.
Karnataka CET-2001
Dual nature of radiation and Matter
142379
The radius of a nucleus with atomic mass number 7 is 2 fermi. The radius of nucleus with atomic mass number 189 is
1 3 fermi
2 4 fermi
3 5 fermi
4 6 fermi
Explanation:
D Given that, $A_{1}=7, A_{2}=189, R_{1}=2$ fermi, $R_{2}$ $=$ ? The expression for nuclear radius is - Then, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{2}{\mathrm{R}_{2}}=\left(\frac{7}{189}\right)^{1 / 3} \Rightarrow \frac{2}{\mathrm{R}_{2}}=\left(\frac{1}{27}\right)^{1 / 3}$ $\frac{2}{\mathrm{R}_{2}}=\frac{1}{3}$ $\mathrm{R}=6 \text { Fermi }$ So,
J and K CET- 2007
Dual nature of radiation and Matter
142384
The X-rays spectrum coming from an X-ray tube
1 no minimum or maximum wavelengths
2 has all wavelength smaller than a certain maximum wavelength
3 has all wavelengths greater than a certain minimum wavelength
4 is monochromatic
Explanation:
C The X-rays spectrum coming from on X-ray tube has all wavelength greater than a certain minimum wavelength. X-rays are electromagnetic waves with wavelength range of $0.1 \AA-100 \AA$. $\mathrm{X}$-ray coming from the tube contains more than one wavelength therefore it is not monochromatic. Let the potential difference is V. Maximum energy of photon, is given by $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}=\mathrm{h} v_{\text {max }}=\frac{\mathrm{hc}}{\lambda_{\text {max }}}$ $\lambda=\frac{\mathrm{hc}}{\mathrm{eV}}$ X-ray beam has energy less than or equal to $E_{\max }$. Hence, wavelength greater than a certain minimum wavelength.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Dual nature of radiation and Matter
142373
The momentum of electrons having a wavelength $2 \AA$ (Given $h=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{m}=$ $9.1 \times 10^{-35} \mathrm{~kg}$ ) is
B Given that, Wavelength $(\lambda)=2 \AA=2 \times 10^{-10} \mathrm{~m}$ Planck's constant $(h)=6.626 \times 10^{-34} \mathrm{Js}$ Let $\mathrm{p}$ be the momentum of the particle. We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{6.62 \times 10^{-34}}{2 \times 10^{-10}}$ $\mathrm{p}=3.31 \times 10^{-24} \mathrm{~kg}^{-\mathrm{ms}^{-1}}$
COMEDK 2015
Dual nature of radiation and Matter
142378
Newton postulated his corpuscular theory on the basis of :
1 Newton's rings
2 colours of thin films
3 rectilinear propagation of light
4 dispersion white light
Explanation:
C Newton postulated his corpuscular theory on the basic of following postulates are- (1) Newton proposed that a source of light emits many minutes particles, elastic, rigid and mass less particles called corpuscles. (2) The particles travel through transparent medium at very high speed in all direction along a straight line. (3) The corpuscles enter our eyes and produces the sensation of vision. (4) Due to different size of the corpuscles they produce different colours. (5) These light particles are replaced by a reflecting surface and attracted by transparent materials. Hence, it explain the rectilinear propagation of light.
Karnataka CET-2001
Dual nature of radiation and Matter
142379
The radius of a nucleus with atomic mass number 7 is 2 fermi. The radius of nucleus with atomic mass number 189 is
1 3 fermi
2 4 fermi
3 5 fermi
4 6 fermi
Explanation:
D Given that, $A_{1}=7, A_{2}=189, R_{1}=2$ fermi, $R_{2}$ $=$ ? The expression for nuclear radius is - Then, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{2}{\mathrm{R}_{2}}=\left(\frac{7}{189}\right)^{1 / 3} \Rightarrow \frac{2}{\mathrm{R}_{2}}=\left(\frac{1}{27}\right)^{1 / 3}$ $\frac{2}{\mathrm{R}_{2}}=\frac{1}{3}$ $\mathrm{R}=6 \text { Fermi }$ So,
J and K CET- 2007
Dual nature of radiation and Matter
142384
The X-rays spectrum coming from an X-ray tube
1 no minimum or maximum wavelengths
2 has all wavelength smaller than a certain maximum wavelength
3 has all wavelengths greater than a certain minimum wavelength
4 is monochromatic
Explanation:
C The X-rays spectrum coming from on X-ray tube has all wavelength greater than a certain minimum wavelength. X-rays are electromagnetic waves with wavelength range of $0.1 \AA-100 \AA$. $\mathrm{X}$-ray coming from the tube contains more than one wavelength therefore it is not monochromatic. Let the potential difference is V. Maximum energy of photon, is given by $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}=\mathrm{h} v_{\text {max }}=\frac{\mathrm{hc}}{\lambda_{\text {max }}}$ $\lambda=\frac{\mathrm{hc}}{\mathrm{eV}}$ X-ray beam has energy less than or equal to $E_{\max }$. Hence, wavelength greater than a certain minimum wavelength.