Question

FIGURE Q39.1 shows the probability density for photons to be detected on the$x$-axis.

a. Is a photon more likely to be detected at $x=0\text{m}$or at $x=$$1\text{m}$ ? Explain.

b. One million photons are detected. What is the expected number of photons in a $1-\text{mm}$-wide interval at $x=0.50\text{m}$?

Short answer

a. The probability density at wide $=1\times {10}^{-3}{\text{m}}^2$.

The expected number of detected photon in the interval of at $x=0.5$is$1\times {10}^3$.

Step-by-step solution

Step.1

The early ${20}^{\text{th}}$century saw the arrival of new kind of physics. The classical deterministic approach was replaced by the probabilistic approach to describe the behavior of really tiny constitution of matter. These particles are called quantum particles and the mechanics that describes the physical behavior of these particles is called quantum physics.

In a quantum world the trajectory a particle could not be exactly predicted. The probability ( $P\left(x\right)$) of finding the particle in a certain location is determined by calculating the expectation value of the wave function $\left(\psi \right(x\left)\right)$of the particle:

$P\left(x\right)=\left|\psi \right(x){|}^2$

Here we will examine the probability of finding the particle in a certain state by looking at the probability density.

The graph below represents the probability density of the photon to be detected on the x -axis.

Part.a.

The total count of photon detected is 1 million. From the graph we can deduce the probability at $x=0.5$is $P(x=0.5)=1$. The probability density at $1\text{mm}$wide interval of $x$at $P(x=0.5)=1$is the area of the curve $P(x=0.5)\Delta x$.

$\begin{array}{l}P(x=0.5)\Delta x=1\left(1\times {10}^{-3}\right)\\ =1\times {10}^{-3}{\text{m}}^2\end{array}$

Part.b.

The expected number of photon that can be detected in the calculated area is

$\begin{array}{l}N=\left(\text{numberofphoton}\right)\text{(areaunderthecurve)}\\ =\left(1\times {10}^6\text{photon}\right)\left(1\times {10}^{-3}\right)\\ =1\times {10}^3\text{photon}\end{array}$

The expected number of detected photon in the interval of$1\text{mm}$ at $x=0.5$is$1\times {10}^3$.