Advertisements
Advertisements
प्रश्न
The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is `v_e = c/100`. Then ______.
- `E_e/E_p = 10^-4`
- `E_e/E_p = 10^-2`
- `p_e/(m_ec) = 10^-2`
- `p_e/(m_ec) = 10^-4`
विकल्प
b and c
a and c
c and d
b and d
उत्तर
b and c
Explanation:
De-Broglie wavelength associated with the charged particles: The energy of a charged particle accelerated through potential difference V is `E = 1/2 mv^2 = qV`
Hence de-Broglie wavelength `λ = h/p = h/sqrt(2mE) = h/sqrt(2mqV)`
`λ_("Electron") = 12.27/sqrt(V) Å, λ_("Proton") = 0.286/sqrt(V) Å`
`λ_("Deutron") = 0.202/sqrt(V) Å, λ_(α"-particle") = 0.101/sqrt(V) Å`
De-Broglie wavelength associated with uncharged particles: For Neutron de-Broglie wavelength is given as
`λ_("Neutron") = (0.286 xx 10^-10)/sqrt(E ("in" e V)) m = 0.286/sqrt(E("in" eV)) Å`
Energy of thermal neutrons at ordinary temperature
∵ E = kT ⇒ λ = `h/sqrt(2mkT)`; where T = Absolute temperature,
k = Boltzman's constant = 1.38 × 10–23 Joule/kelvin
So, `λ_("Thermal neutron") = (6.62 xx 10^-34)/sqrt(2 xx 1.67 xx 10^-27 xx 1.38 xx 10-23 T) = 30.83/sqrt(T) Å`
Mass of electron = me, Mass of photon = mp,
Velocity of electron is equal to velocity of proton which are ve and vp respectively.
De-Broglie wavelength for electron, `λ_e h/sqrt(m_e v_e)`
⇒ `λ_e = h/(m_e(c/100)) = (100 h)/(m_e c)` .....(i)
Kinetic energy of electron, `E_e = 1/2 m_e v_e^2`
⇒ `m_e v_e = sqrt(2E_e m_e)`
So, `λ_e = h/(m_e v_e) = h/sqrt(2m_e E_e)`
⇒ `E_e = h^2/(2λ_e^2 m_e)` .....(ii)
The wavelength of proton is λp and having an energy is
∴ `E_p = (hc)/λ_p = (hc)/(2λ_e)` ......[∵ λp = 2λe]
∴ `E_p/E_e = ((hc)/(2λ_e)) ((2λ_e^2 m_e)/h^2)`
= `(λ_em_ec)/h = (100h)/(m_ec) xx (m_ec)/h` = 100
So, `E_e/E_p = 1/100 = 10^-2`
For electron, `p_e = m_ev_e = m_e xx c/100`
So, `p_e/(m_ec) = 1/100 = 10^-2`
Important point:
Ratio of the wavelength of photon and electron: The wavelength of a photon of energy E is given by `λ_(ph) = (hc)/E`
While the wavelength of an electron of kinetic energy K is given by `λ_c = h/sqrt(2mK)`
Therefore for the same energy the ratio `λ_(ph)/λ_e = c/E sqrt(2mK) = sqrt((2mc^2 K)/E^2`
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?
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.
Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
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 ______.
Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle (∆x∆p ≃ h). You can assume the uncertainty in position ∆x as 1 nm. Assuming p ≃ ∆p, find the energy of the electron in electron volts.
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).
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
