<span>d. electron
J J Thomson discovered the electron, and it was put in his model of the atom.</span>
Answer:
a) (0, -33, 12)
b) area of the triangle : 17.55 units of area
Explanation:
<h2>
a) </h2>
We know that the cross product of linearly independent vectors 
and 
gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.
Luckily for us, we know that vectors 
and 
are living in the plane through the points P, Q, and R, and are linearly independent.
We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).
If they weren't linearly independent, we will obtain vector zero as the result of the cross product.
So, for our problem:
<h2>B)</h2>
We know that and are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:
so:
Answer:
The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.
Explanation:
According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.
σx . σpx ≥ h/4π
where,
h is the Planck´s constant
If σx = 5 × 10⁻¹²m,
5 × 10⁻¹²m . σpx ≥ 6.63 × 10⁻³⁴ kg.m².s⁻¹/4π
σpx ≥ 1 × 10⁻²³ kg.m.s⁻¹
