An electron travelling at 3.7 x 105 m/s has an uncertainty in its velocity of 1.88 x 105 m/s. What is the uncertainty in its position (in pm)? (Just put the number, don't use scientific notation and decimal point in the final answer) Avogadro's number = 6.022 x 1023, h = 6.626 x 10-34 J. s, c = 3.00 x 108 m/s, me= 9.109 x 10-31 kg

The uncertainty in its position is 308 pm

Explanation:

Given;

velocity of electron, v = 3.7 x 10⁵ m/s

uncertainty in velocity of the electron, Δv =   1.88 x 10⁵ m/s

Planck's constant, h = 6.626 x 10⁻³⁴ J.s

mass of electron, m = 9.109 x 10⁻³¹ kg

Using Heisenberg's uncertainty principle, we will determine the uncertainty in the position of the electron.

Δx(mΔv) ≥ h/4π

where;

Δx is uncertainty in position

m is mass of the electron

Δv is uncertainty in velocity of the electron

h is Planck's constant

Δx

Δx = 308 x 10⁻¹² m

Δx = 308 pm

Therefore,  the uncertainty in its position is 308 pm

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)