Question

uncertainty in position of an electron is 1 nm. calculate its uncertainty in velocity (given h=6.626 into 10th the power -34 js)

Answer 1

Answer:

We have the relation,

(∆ x) /(∆ v) = h/4πm......... 1

Here, h= 6.626x 10^-34 JS= 6.626x 10^-27 erg sec, ∆ x= 1 nm= 1x 10^-7 cm

We know,

m= 9.1x 10^-28 g

∆ v= 6.626x 10^-27 erg sec/4x 3.142x 9.1x 10^-28 gx 1x 10^-7 CM

= 6.62x 10^-27 /114.37x 10^-35 CM/sec

= 0.579x 10^ 7 CM/ sec....

Thank you

Answer 2

Answer:

Explanation:

We know according to the Heisenberg uncertainty principle.

∆X*∆P = hnπWhere ∆X = uncertainty in position and ∆P is the uncertainty in the position.Where ∆P = m∆V ( m = mass and ∆V = uncertainty in velocity)So,∆X*m∆V = hnπFor electron n = 4∆X = 1 nm = 1*10−9 mMass of electron = 9.1*10−31 Kgh= 6.626*10−34 Jsπ = 3.14So,∆V = h4πm*∆X =6.626*10−344*3.14*9.1*10−31*10−9