Advertisements
Advertisements
प्रश्न
Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn= 1.675 × 10−27 kg)
Advertisements
उत्तर
De Broglie wavelength = 2.327 × 10−12 m; neutron is not suitable for the diffraction experiment.
Kinetic energy of the neutron, K = 150 eV
= 150 × 1.6 × 10−19
= 2.4 × 10−17 J
Mass of a neutron, mn = 1.675 × 10−27 kg
The kinetic energy of the neutron is given by the relation:
`"K" = 1/2 "m"_"n""v"^2`
`"m"_"n""v" = sqrt(2 "KM"_"n")`
Where
v = Velocity of the neutron
mnv = Momentum of the neutron
De-Broglie wavelength of the neutron is given as:
`lambda = "h"/("m"_"n""v") = "h"/(sqrt(2 "Km"_"n"))`
it is clear that wavelength is inversely proportional to the square rot of mass.
Hence wavelength decrease with an increase in mass and vice versa.
∴ `lambda = (6.6 xx 10^(-34))/sqrt(2 xx 2.4 xx 10^(-17) xx 1.675 xx 10^(-27))`
= 2.327 × 10−12 m
It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, i.e., 10−10 m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.
APPEARS IN
संबंधित प्रश्न
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of `(3/2)` kT at 300 K.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
A electron of mass me revolves around a nucleus of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated wit it is expressed as `vecμ =−e/(2 m_e)vecL `, where `vec L` is the orbital angular momentum of the electron. Give the significance of negative sign.
What are matter waves?
Show with the help of a labelled graph how their wavelength (λ) varies with their linear momentum (p).
70 cal of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the gas through the same range at constant volume will be (assume R = 2 cal/mol-K).
An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then ______.
A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to ______.
A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values λ1, λ2 with λ1 > λ2. Which of the following statement are true?
- The particle could be moving in a circular orbit with origin as centre.
- The particle could be moving in an elliptic orbit with origin as its focus.
- When the de Broglie wavelength is λ1, the particle is nearer the origin than when its value is λ2.
- When the de Broglie wavelength is λ2, the particle is nearer the origin than when its value is λ1.
Two particles move at a right angle to each other. Their de-Broglie wavelengths are λ1 and λ2 respectively. The particles suffer a perfectly inelastic collision. The de-Broglie wavelength λ, of the final particle, is given by ______.
The De-Broglie wavelength of electron in the third Bohr orbit of hydrogen is ______ × 10-11 m (given radius of first Bohr orbit is 5.3 × 10-11 m):
How will the de-Broglie wavelength associated with an electron be affected when the accelerating potential is increased? Justify your answer.
The graph which shows the variation of `(1/lambda^2)` and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
