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.
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
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.
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 ______.
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 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 ______.
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 ______.
Two particles A1 sand A2 of masses m1, m2 (m1 > m2) have the same de Broglie wavelength. Then ______.
- their momenta are the same.
- their energies are the same.
- energy of A1 is less than the energy of A2.
- energy of A1 is more than the energy of A2.
A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values λ1, λ2 with λ1 > λ2. Which of the following statement are true?
- The particle could be moving in a circular orbit with origin as centre.
- The particle could be moving in an elliptic orbit with origin as its focus.
- When the de Broglie wavelength is λ1, the particle is nearer the origin than when its value is λ2.
- When the de Broglie wavelength is λ2, the particle is nearer the origin than when its value is λ1.
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).
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)
An electron is accelerated from rest through a potential difference of 100 V. Find:
- the wavelength associated with
- the momentum and
- the velocity required by the electron.
The De-Broglie wavelength of electron in the third Bohr orbit of hydrogen is ______ × 10-11 m (given radius of first Bohr orbit is 5.3 × 10-11 m):
Matter waves are ______.
