Advertisements
Advertisements
प्रश्न
A proton and an α-particle have the same de-Broglie wavelength Determine the ratio of their speeds.
Advertisements
उत्तर
The de Broglie wavelength (λ) of a particle is also given by
`lambda=h/(mv)`
Here,
h = Planck's constant
m = Mass
v = Speed of the particle
`:.lambda"proton"=h/(m_"proton"v_"proton")`
`lambda"alpha"=h/(_"alpha"v_"alpha")`
λproton=λalpha
`therefore h/(m_"proton"v_"proton")=h/(4m_"proton"v_"alpha")`
`therefore (4h)/h=(m_"proton"v_"proton")/(m_"proton"v_"alpha")`
`therefore 4=(v_"proton")/(v_"alpha")`
`=>(v_"proton")/v_"alpha"=4`
APPEARS IN
संबंधित प्रश्न
For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?
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.
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?
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
Show with the help of a labelled graph how their wavelength (λ) varies with their linear momentum (p).
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.
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`
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)
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:
How will the de-Broglie wavelength associated with an electron be affected when the accelerating potential is increased? Justify your answer.
