The uncertainty in the position of an electron along an x axis is given as 50 pm

Question

The uncertainty in the position of an electron along an x axis is given as 50 pm, which is given about equal to the radius of a hydrogen atom. What is the least uncertainty in any simultaneous measurement of the momentum component rho _x of this electron? Consider a potential energy barrier whose height U_b is 6.0 eV and whose thickness L is 0.70 nm. What is the energy of an incident electron whose transmission coefficient is 0.0010? A proton is confined to a one-dimensional infinite potential well 100 pm wide. What is its ground-state energy?

Explanation / Answer

Uncertainty in position of electron (delta x)= 50 pm = 50* 10^-12 m

Let Uncertainty in momentum of electron is =(delta p)

According to Heisenberg uncertainty principle-

                        (delta x)(delta p) =( h-bar)/2            where h-bar=h/2, h is Planck’s constant.

So        delta p = 1.054*10^-34 /(2*50*10^-12)

                        = 1.054 * 10^-24 Kg-m/s.

Uncertainty in momentum of electron is= 1.054 * 10^-24 Kg-m/s.

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