Advertisements
Advertisements
प्रश्न
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Advertisements
उत्तर
Kinetic energy of the electron, Ek = 120 eV
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
(a) For the electron, we can write the relation for kinetic energy as:
`"E"_"k" = 1/2"mv"^2`
Where,
v = Speed of the electron
∴ `"v"^2 = sqrt((2"eE"_"k")/"m")`
= `sqrt((2 xx 1.6 xx 10^(-19) xx 120)/(9.1 xx 10^(-31)))`
= `sqrt(42.198 xx 10^12)`
= 6.496 × 106 m/s
Momentum of the electron, p = mv
= 9.1 × 10−31 × 6.496 × 106
= 5.91 × 10−24 kg m s−1
Therefore, the momentum of the electron is 5.91 × 10−24 kg m s−1.
(b) Speed of the electron, v = 6.496 × 106 m/s
(c) De Broglie wavelength of an electron having a momentum p, is given as:
`lambda = "h"/"p"`
= `(6.6 xx 10^(-34))/(5.91 xx 10^(-24))`
= 1.116 × 10−10 m
= 0.112 nm
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
APPEARS IN
संबंधित प्रश्न
For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?
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.
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.
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?
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.
State any one phenomenon in which moving particles exhibit wave nature.
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
The de-Broglie wavelength associated with a material particle when it is accelerated through a potential difference of 150 volt is 1 Å. What will be the de-broglie wavelength associated with the same particle when it is accelerated through a potential difference of 4500 V?
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 proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.
An electron (mass m) with an initial velocity `v = v_0hati (v_0 > 0)` is in an electric field `E = - E_0hati `(E0 = constant > 0). It’s de Broglie wavelength at time t is given by ______.
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 ______.
Two particles A and B of de Broglie wavelengths λ1 and λ2 combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).
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 ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : `sqrt2`. Then, the ratio of Vp to Vd will be ______.
An electron of mass me, and a proton of mass mp = 1836 me are moving with the same speed. The ratio of the de Broglie wavelength `lambda_"electron"/lambda_"proton"` will be:
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):
