Advertisements
Advertisements
Question
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`
Options
b and c
a and c
c and d
b and d
Advertisements
Solution
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
RELATED QUESTIONS
Describe the construction of photoelectric cell.
A proton and an α-particle have the same de-Broglie wavelength Determine the ratio of their speeds.
Calculate the momentum of the electrons accelerated through a potential difference of 56 V.
What is the de Broglie wavelength of a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s?
What is the de Broglie wavelength of a ball of mass 0.060 kg moving at a speed of 1.0 m/s?
What is the de Broglie wavelength of a dust particle of mass 1.0 × 10−9 kg drifting with a speed of 2.2 m/s?
For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?
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)
Describe briefly how the Davisson-Germer experiment demonstrated the wave nature of electrons.
A proton and α-particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength λp to that λα is _______.
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).
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.
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.
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 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):
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 ______.
Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength?
How will the de-Broglie wavelength associated with an electron be affected when the accelerating potential is increased? Justify your answer.
Matter waves are ______.
