If uncertainity in position is zero, the uncertainity in momentum of a

1.	If uncertainity in position is zero, the uncertainity in momentum of an electron will be
Answer» `1/m sqrt(h/(4LAMBDA))` `msqrt(h)/(sqrt(4lambda))` `sqrt((h)/(4lambdam))` `sqrt((hm)/(4lambda))` Solution :`Delta x . ""m Delta v GE h/(4pi)` and `LAMBDA = h/(mv)` In the present case `Delta x = Delta v ` `therefore m (Deltav)^2 ge h/(4pi) IMPLIES m^2 (Deltav)^2 ge (mh)/(4pi)` `impliesm.Delta x ge sqrt((mh)/(4pi))`

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