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
Advertisements
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
Calculate the momentum of the electrons accelerated through a potential difference of 56 V.
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
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.
Describe briefly how the Davisson-Germer experiment demonstrated the wave nature of electrons.
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 ______.
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 ______.
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 ______.
An electron is moving with an initial velocity `v = v_0hati` and is in a magnetic field `B = B_0hatj`. Then it’s de Broglie wavelength ______.
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.
An alpha particle is accelerated through a potential difference of 100 V. Calculate:
- The speed acquired by the alpha particle, and
- The de-Broglie wavelength is associated with it.
(Take mass of alpha particle = 6.4 × 10−27 kg)
For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?
E, c and `v` represent the energy, velocity and frequency of a photon. Which of the following represents its wavelength?
