Advertisements
Advertisements
Question
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.
Advertisements
Solution
Wavelength of light of a sodium line, λ = 589 nm = 589 × 10−9 m
Mass of an electron, me = 9.1 × 10−31 kg
Mass of a neutron, mn = 1.66 × 10−27 kg
Planck’s constant, h = 6.6 × 10−34 Js
(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
`"K" = 1/2 "m"_"e""v"^2` ...............(1)
We have the relation for de Broglie wavelength as:
`lambda = "h"/("m"_"e""v")`
∴ `"v"^2 = "h"^2/(lambda^2"m"_"e"^2)` ........(2)
Substituting equation (2) in equation (1), we get the relation:
`"K" = 1/2 ("m"_"e""h"^2)/(lambda^2"m"_"e"^2) = "h"^2/(2lambda^2"m"_"e")` ..........(3)
= `(6.6 xx 10^(-34))^2/(2 xx (589 xx 10^(-9))^2 xx 9.1 xx 10^(-31))`
≈ 6.9 × 10−25 J
= `(6.9 xx 10^(-25))/(1.6 xx 10^(-19))`
= 4.31 × 10−6 eV
= 4.31 μeV
Hence, the kinetic energy of the electron is 6.9 × 10−25 J or 4.31 μeV.
(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
`"h"^2/(2lambda^2 "m"_"n")`
= `(6.6 xx 10^(-34))^2/(2 xx (589 xx 10^(-9))^2 xx 1.66 xx 10^(-27))`
= 3.78 × 10−28 J
= `(3.78 xx 10^(-28))/(1.6 xx 10^(-19))`
= 2.36 × 10−9 eV
= 2.36 neV
Hence, the kinetic energy of the neutron is 3.78 × 10−28 J or 2.36 neV.
APPEARS IN
RELATED QUESTIONS
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
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)
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.
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.
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).
The wavelength of the matter wave is dependent on ______.
A proton and α-particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength λp to that λα is _______.
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 (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 ______.
A particle A with a mass m A is moving with a velocity v and hits a particle B (mass mB) at rest (one dimensional motion). Find the change in the de Broglie wavelength of the particle A. Treat the collision as elastic.
Given below are two statements:
Statement - I: Two photons having equal linear momenta have equal wavelengths.
Statement - II: If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.
What is another name for matter waves?
In \[\lambda=\frac{h}{mv}\], what do \[m\] and \[v\] represent, respectively?
Which wave quantities did de Broglie relate to the particle quantities energy and momentum?
What value of Planck's constant is used in the de Broglie wavelength calculations?
