Advertisements
Advertisements
प्रश्न
The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
E = hv, p = `"h"/lambda`
But while the value of λ is physically significant, the value of v (and therefore, the value of the phase speed vλ) has no physical significance. Why?
Advertisements
उत्तर
The absolute value of the energy of a particle is arbitrary within the additive constant. Hence, wavelength (λ) is significant, but the frequency (v) associated with an electron has no direct physical significance.
Therefore, the product νλ(phase speed)has no physical significance.
Group speed is given as:
`"v"_"G" = ("dv")/("dk")`
= `"dv"/("d"(1/lambda)) = "dE"/"dp" = ("d"("p"^2/2"m"))/"dp" = "p"/"m"`
This quantity has a physical meaning.
APPEARS IN
संबंधित प्रश्न
What is the de Broglie wavelength of a ball of mass 0.060 kg moving at a speed of 1.0 m/s?
What is the de Broglie wavelength of a dust particle of mass 1.0 × 10−9 kg drifting with a speed of 2.2 m/s?
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me = 9.11 × 10−31 kg).
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)
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.
The wavelength λ of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of a photon in (2λmc/h) times the kinetic energy of electron; where m, c and h have their usual meaning.
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
Sodium and copper have work function 2.3 eV and 4.5 eV respectively. Then, the ratio of the wavelengths is nearest to ______.
Which one of the following deflect in electric field
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 ______.
An electron (mass m) with an initial velocity `v = v_0hati` is in an electric field `E = E_0hatj`. If λ0 = h/mv0, it’s de Broglie wavelength at time t is given by ______.
Relativistic corrections become necessary when the expression for the kinetic energy `1/2 mv^2`, becomes comparable with mc2, where m is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?
- λ = 10 nm
- λ = 10–1 nm
- λ = 10–4 nm
- λ = 10–6 nm
The equation λ = `1.227/"x"` nm can be used to find the de Brogli wavelength of an electron. In this equation x stands for:
Where,
m = mass of electron
P = momentum of electron
K = Kinetic energy of electron
V = Accelerating potential in volts for electron
A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be:
Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength?
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):
