Advertisements
Advertisements
Question
A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp and λa related to each other?
Advertisements
Solution
In this problem since both proton and α-particle are accelerated through the same potential difference,
We know that, `λ = h/sqrt(2mqv)`
∴ `λ oo 1/sqrt(mq)`
`λ_p/λ_α = sqrt(m_αq_α)/(m_pq_P) = sqrt(4m_p xx 2e)/sqrt(m_p xx e) = sqrt(8)`
∴ `λ_p = sqrt(8)λ_α`
i.e., wavelength of proton is `sqrt(8)` times wavelength of α-particle.
Important point: 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_(a"-particle") = 0.101/sqrt(V) Å`
APPEARS IN
RELATED QUESTIONS
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
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.
What is the de Broglie wavelength of a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s?
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
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.
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.
Describe briefly how the Davisson-Germer experiment demonstrated the wave nature of electrons.
The de-Broglie wavelength associated with a material particle when it is accelerated through a potential difference of 150 volt is 1 Å. What will be the de-broglie wavelength associated with the same particle when it is accelerated through a potential difference of 4500 V?
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 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 ______.
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`
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.
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.
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 ______.
A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be:
In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to ______.
How will the de-Broglie wavelength associated with an electron be affected when the velocity of the electron decreases? Justify your answer.
