Question

When 5 X 10¹² photons pass through an experimental apparatus, 2.0 X 10⁹ land in a 0.10-mm-wide strip. What is the probability density at this point?

Short answer

Thus, the probability density is$4m^{-1}$

Step-by-step solution

Given information

When 5 X 10¹² photons pass through an experimental apparatus, 2.0 X 10⁹ land in a 0.10-mm-wide strip.

Explanation

The probability will be:

$$\begin{gathered} P=\frac{2.0\times {10}^9}{5\times {10}^{12}} \\ P=0.0004 \end{gathered}$$

Therefore, the probability density is:

$$\begin{gathered} P\left(x\right)=\frac{P}{\delta x} \\ P\left(x\right)=\frac{0.0004}{0.10\times {10}^{-3}} \\ P\left(x\right)=4{\mathrm{m}}^{-1} \end{gathered}$$