307233
At temperature T, the average kinetic energy of any particle is \(\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}{\rm{.}}\) The de Broglie wavelength follows the order:
1 Visible photon > Thermal neutron > Thermal electron
2 Thermal proton > Thermal electron > Visible photon
3 Thermal proton > Visible photon > Thermal electron
4 Visible photon > Thermal electron > Thermal neutron
Explanation:
Kinetic energy of any particle \({\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}\) Also, \({\rm{K}}{\rm{.E}}{\rm{. = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}\) \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}} \Rightarrow {{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3kT}}}}{{\rm{m}}}\) \({\rm{v = }}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} \) De-broglie wavelength \({\rm{ = \lambda = }}\frac{{\rm{h}}}{{{\rm{mv}}}}{\rm{ = }}\frac{{\rm{h}}}{{{\rm{m}}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} }}\) \({\rm{\lambda = }}\frac{{\rm{h}}}{{\sqrt {{\rm{3kTm}}} }}\;,\;{\rm{\lambda }}\, \propto \frac{{\rm{1}}}{{\sqrt {\rm{m}} }}\) Mass of electron < mass of neutron \({\rm{\lambda }}\) (electron) > \({\rm{\lambda }}\) (neutron)
JEE - 2015
CHXI02:STRUCTURE OF ATOM
307235
If the Planck’s constant \({\rm{h = 6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{Js}}\), the de-Broglie wavelength of a particle having momentum of \({\rm{3}}{\rm{.3 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}{\rm{kg}}{\rm{.m}}{\rm{.}}{{\rm{s}}^{{\rm{ - 1}}}}\) will be
307237
Two particles A and B are in motion. If the wavelength associated with the particle A is \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{m,}}\) calculate the wavelength of particle B if its momentum is half of A.
307233
At temperature T, the average kinetic energy of any particle is \(\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}{\rm{.}}\) The de Broglie wavelength follows the order:
1 Visible photon > Thermal neutron > Thermal electron
2 Thermal proton > Thermal electron > Visible photon
3 Thermal proton > Visible photon > Thermal electron
4 Visible photon > Thermal electron > Thermal neutron
Explanation:
Kinetic energy of any particle \({\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}\) Also, \({\rm{K}}{\rm{.E}}{\rm{. = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}\) \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}} \Rightarrow {{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3kT}}}}{{\rm{m}}}\) \({\rm{v = }}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} \) De-broglie wavelength \({\rm{ = \lambda = }}\frac{{\rm{h}}}{{{\rm{mv}}}}{\rm{ = }}\frac{{\rm{h}}}{{{\rm{m}}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} }}\) \({\rm{\lambda = }}\frac{{\rm{h}}}{{\sqrt {{\rm{3kTm}}} }}\;,\;{\rm{\lambda }}\, \propto \frac{{\rm{1}}}{{\sqrt {\rm{m}} }}\) Mass of electron < mass of neutron \({\rm{\lambda }}\) (electron) > \({\rm{\lambda }}\) (neutron)
JEE - 2015
CHXI02:STRUCTURE OF ATOM
307235
If the Planck’s constant \({\rm{h = 6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{Js}}\), the de-Broglie wavelength of a particle having momentum of \({\rm{3}}{\rm{.3 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}{\rm{kg}}{\rm{.m}}{\rm{.}}{{\rm{s}}^{{\rm{ - 1}}}}\) will be
307237
Two particles A and B are in motion. If the wavelength associated with the particle A is \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{m,}}\) calculate the wavelength of particle B if its momentum is half of A.
307233
At temperature T, the average kinetic energy of any particle is \(\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}{\rm{.}}\) The de Broglie wavelength follows the order:
1 Visible photon > Thermal neutron > Thermal electron
2 Thermal proton > Thermal electron > Visible photon
3 Thermal proton > Visible photon > Thermal electron
4 Visible photon > Thermal electron > Thermal neutron
Explanation:
Kinetic energy of any particle \({\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}\) Also, \({\rm{K}}{\rm{.E}}{\rm{. = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}\) \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}} \Rightarrow {{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3kT}}}}{{\rm{m}}}\) \({\rm{v = }}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} \) De-broglie wavelength \({\rm{ = \lambda = }}\frac{{\rm{h}}}{{{\rm{mv}}}}{\rm{ = }}\frac{{\rm{h}}}{{{\rm{m}}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} }}\) \({\rm{\lambda = }}\frac{{\rm{h}}}{{\sqrt {{\rm{3kTm}}} }}\;,\;{\rm{\lambda }}\, \propto \frac{{\rm{1}}}{{\sqrt {\rm{m}} }}\) Mass of electron < mass of neutron \({\rm{\lambda }}\) (electron) > \({\rm{\lambda }}\) (neutron)
JEE - 2015
CHXI02:STRUCTURE OF ATOM
307235
If the Planck’s constant \({\rm{h = 6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{Js}}\), the de-Broglie wavelength of a particle having momentum of \({\rm{3}}{\rm{.3 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}{\rm{kg}}{\rm{.m}}{\rm{.}}{{\rm{s}}^{{\rm{ - 1}}}}\) will be
307237
Two particles A and B are in motion. If the wavelength associated with the particle A is \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{m,}}\) calculate the wavelength of particle B if its momentum is half of A.
307233
At temperature T, the average kinetic energy of any particle is \(\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}{\rm{.}}\) The de Broglie wavelength follows the order:
1 Visible photon > Thermal neutron > Thermal electron
2 Thermal proton > Thermal electron > Visible photon
3 Thermal proton > Visible photon > Thermal electron
4 Visible photon > Thermal electron > Thermal neutron
Explanation:
Kinetic energy of any particle \({\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}\) Also, \({\rm{K}}{\rm{.E}}{\rm{. = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}\) \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}} \Rightarrow {{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3kT}}}}{{\rm{m}}}\) \({\rm{v = }}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} \) De-broglie wavelength \({\rm{ = \lambda = }}\frac{{\rm{h}}}{{{\rm{mv}}}}{\rm{ = }}\frac{{\rm{h}}}{{{\rm{m}}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} }}\) \({\rm{\lambda = }}\frac{{\rm{h}}}{{\sqrt {{\rm{3kTm}}} }}\;,\;{\rm{\lambda }}\, \propto \frac{{\rm{1}}}{{\sqrt {\rm{m}} }}\) Mass of electron < mass of neutron \({\rm{\lambda }}\) (electron) > \({\rm{\lambda }}\) (neutron)
JEE - 2015
CHXI02:STRUCTURE OF ATOM
307235
If the Planck’s constant \({\rm{h = 6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{Js}}\), the de-Broglie wavelength of a particle having momentum of \({\rm{3}}{\rm{.3 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}{\rm{kg}}{\rm{.m}}{\rm{.}}{{\rm{s}}^{{\rm{ - 1}}}}\) will be
307237
Two particles A and B are in motion. If the wavelength associated with the particle A is \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{m,}}\) calculate the wavelength of particle B if its momentum is half of A.
307233
At temperature T, the average kinetic energy of any particle is \(\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}{\rm{.}}\) The de Broglie wavelength follows the order:
1 Visible photon > Thermal neutron > Thermal electron
2 Thermal proton > Thermal electron > Visible photon
3 Thermal proton > Visible photon > Thermal electron
4 Visible photon > Thermal electron > Thermal neutron
Explanation:
Kinetic energy of any particle \({\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}}\) Also, \({\rm{K}}{\rm{.E}}{\rm{. = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}\) \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{kT}} \Rightarrow {{\rm{v}}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3kT}}}}{{\rm{m}}}\) \({\rm{v = }}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} \) De-broglie wavelength \({\rm{ = \lambda = }}\frac{{\rm{h}}}{{{\rm{mv}}}}{\rm{ = }}\frac{{\rm{h}}}{{{\rm{m}}\sqrt {\frac{{{\rm{3kT}}}}{{\rm{m}}}} }}\) \({\rm{\lambda = }}\frac{{\rm{h}}}{{\sqrt {{\rm{3kTm}}} }}\;,\;{\rm{\lambda }}\, \propto \frac{{\rm{1}}}{{\sqrt {\rm{m}} }}\) Mass of electron < mass of neutron \({\rm{\lambda }}\) (electron) > \({\rm{\lambda }}\) (neutron)
JEE - 2015
CHXI02:STRUCTURE OF ATOM
307235
If the Planck’s constant \({\rm{h = 6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{Js}}\), the de-Broglie wavelength of a particle having momentum of \({\rm{3}}{\rm{.3 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}{\rm{kg}}{\rm{.m}}{\rm{.}}{{\rm{s}}^{{\rm{ - 1}}}}\) will be
307237
Two particles A and B are in motion. If the wavelength associated with the particle A is \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{m,}}\) calculate the wavelength of particle B if its momentum is half of A.
