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The French physicist Louis de Broglie in...

The French physicist Louis de Broglie in 1924 postulated that matter like radiation , should exhibit a dual behaviour. He proposed the following relationship between the wavelength `.lambda` of a material particle,its linear momentum P and Planck constant h.
`lambda=(h)/(p)=(h)/(mv)`
The de Broglie relaion that the wavelength of a particle should decrease as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles.The waves or de Broglie waves. These waves differ from the electromagnetic waves as they:
(i) have lower velocities
(ii) have no electrical and magnetic fields and
(iii) are not emitted by the particle under consideration.
The expermental confirmation of the de Broglie relation was obtained when Davission and Germer in 1927, observed. As diffraction is a characteristic property of waves, hence the beam of electrons behave as a wave as proposed by de Broglie.
Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron . He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we can determine its position. The converse is also true. This is summed up in what we now call the ''Heisenberg uncertainty principle'' : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of uncertainty in the position, `Deltax` and the uncertainty in the momentum `Delta(mv)` must be greater than or equal to `(h)/(4pi), i.e.,`
`Deltax Delta(mv)ge(h)/(4pi)`
The transition so that the de Broglie wavelength of electron becomes 3 times of its initial value in `He^(+)` ion will be :

A

`2 to 5`

B

`3 to 2`

C

`2 to 6`

D

`1 to 2`

Text Solution

Verified by Experts

The correct Answer is:
c
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