I’m 95% sure it’s covalent bonds.
Answer:
1.170*10^-3 m
3.23*10^-32 m
Explanation:
To solve this, we apply Heisenberg's uncertainty principle.
the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"
Δp * Δx = h/4π
m(e).Δv * Δx = h/4π
If we make Δx the subject of formula, by rearranging, we have
Δx = h / 4π * m(e).Δv
on substituting the values, we have
for the electron
Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 5.67*10^-31
Δx = 1.170*10^-3 m
for the bullet
Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 0.021
Δx = 3.23*10^-32 m
therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet
Answer:
ni = 2.04e19
Explanation:
we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have
n = p = ni
from intrinsic carrier concentration



1.7 = ni * 1.6*10^{-19} * (.35 + .17)
ni = 2.014 *10^{19} m^{-3}
ni = 2.04e19
Answer:
W = 100000 J = 100 KJ
Explanation:
Here we will use the most basic and general formula of work, which is as follows:

where,
W = Work Done = ?
F = Force Required = 200 N
d = Length of Track = 500 m
Therefore,

<u>W = 100000 J = 100 KJ</u>
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