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We can pin point an aeroplane moving in the sky, whatever may be its speed i.e., we can locate both its exact position as wellas direction . However, it is not possible to doso in case of a moving microscopic particle such as electron. In fact, we cannot see any such particle without disturbing it. This has been stated by Heisenberg in the form of uncertainty principle. The mathematical form of this principle is `: Deltax.DeltaP ge (h)/(4pi)` (constant). Since the product of `Deltax` and `Deltap(m Delta upsilon)`is constant , if one is very small, the other is bound to be large. The principle as such has no significance in daily life since it applies to those particles which we can not see.
If uncertainty in position and momentum are equal , then the uncertainty in velocity is :

A

`(1)/(2m) sqrt((h)/(pi))`

B

`sqrt((h)/(2pi))`

C

`sqrt((h)/(pi))`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A

`Delta x Deltap=(h)/(4 pi) or Deltap^(2)=(h)/(4pi)`
`Delta (m upsilon)^(2)=(h)/(4pi)`
`(Delta upsilon)^(2)=(1)/(m^(2))xx(h)/(4pi) , Delta upsilon =(1)/(2m)sqrt((h)/(pi))`.
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We can pin point an aeroplane moving in the sky, whatever may be its speed i.e., we can locate both its exact position as wellas direction . However, it is not possible to doso in case of a moving microscopic particle such as electron. In fact, we cannot see any such particle without disturbing it. This has been stated by Heisenberg in the form of uncertainty principle. The mathematical form of this principle is : Deltax.DeltaP ge (h)/(4pi) (constant). Since the product of Deltax and Deltap(m Delta upsilon) is constant , if one is very small, the other is bound to be large. The principle as such has no significance in daily life since it applies to those particles which we can not see. If the uncertainty in the position of electron is zero , the uncertainty in its momentum would be

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