In an atom, an electron is confined to a space of roughly 10−10 meters. if we take this to be the uncertainty in the electron's position, what is the minimum uncertainty δp in its momentum? express your answer in kilogram meters per second to two significant figures.

Δp = 3.31 × 10⁻²⁴ kg.m/s

Explanation:

Given:

Electron is confined to a space of,  Δx  = 10⁻¹⁰ m

According to the Heisenberg uncertainty principle, we have

where, h is the plank's constant = 6.626 × 10⁻³⁴ J.s

Δp is the uncertainty in the momentum

on substituting the respective values, we get

or

Δp = 3.31 × 10⁻²⁴ kg.m/s

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

yes the answer is total mass in unchanged   : )