Yes!
Now find the probability density (probability per unit length) at the position of the 10th atom.

Where l = 11b, n = 3 and x = 9b
y2 = [2/(11X0.15)]sin2[3p(9b)/(11b)]
= 1.21 sin2(27/11) p {angle in radians}
= 1.21 sin2(27/11)180° {angle in degrees}
= 1.21 sin2 442°
= 1.21 X (0.99)2 = 1.19 nm-1

Probability of finding the electron in a distance 0.0050 nm centered at the 10th atom
= 1.19 nm-1 X 0.0050 nm = 5.9X10-3

Return to Self Test 3