Question

consider this probability density in a box of size l = 3.5 nm: (x) = nx for 0 ≤ x ≤ l/2 and = n(l-x) for l/2 ≤ x ≤ l , (x) = 0 outside the box.what is the value of n, the normalization constant?

Answer 1

To find the normalization constant, we need to ensure that the probability density function integrates to 1 over the entire range of x. In this case, the range is from 0 to l. The value of n, the normalization constant, is 8 divided by the square of the box's size (l).

Step-by-step explanation:

To find the normalization constant, we need to ensure that the probability density function integrates to 1 over the entire range of x. In this case, the range is from 0 to l. We can split the integration into two parts:

1. For 0 ≤ x ≤ l/2, the probability density function is nx. So, the integral for this part is ∫(0 to l/2) nx dx = n*l^2/8.
2. For l/2 ≤ x ≤ l, the probability density function is n(l-x). So, the integral for this part is ∫(l/2 to l) n(l-x) dx = n*l^2/8.

Since the sum of the integrals for the two parts should equal 1, we have n*l^2/8 + n*l^2/8 = 1. Solving for n, we get n = 8/l^2.