Question

An electron is allowed to move freely in a closed cubic box of length of side 10 cm. The maximum uncertainty in its velocity will be observed as

Answer 1

Answer:

The answer will be  5 x 10^(-4) m/s

Explanation:

According to the problem the electron is allowed to move freely in a closed box.

Therefore, maximum uncertainty in position will be equal to the length.

let l is the length

∆l= √3 x 10

∆l (m∆v)= h/ 4 π [h = plank's constant, v is the velocity, m is the mass of the electron]

∆v=  h/ 4 π x ∆l (m)

    =  6.623 x 10^(-34)/ 4 x 3.14 x 10√3 x 10^3 x 9.1 x 10^(-31)

   =  5 x 10^(-4) m/s