Question

The probability density of finding a particle somewhere along the$x-axisis0forx61mm.Atx=1mm,$the probability density is $c$. For$xU1mm$, the probability density decreases by a factor of $8$each time the distance from the origin is doubled. What is the probability that the particle will be found in the interval $2mm...x...4mm?$

Short answer

$p\left(x\right)isprobabilitydensity$

Step-by-step solution

Probability of finding a particle in certain interval δx is given by

probability to finding a particle in $a\le x\le b={\int }_a^{b}p\left(x\right)dx$

Calculate the value C by using normalization condition,

$$\begin{gathered} {\int }_{-\infty }^{+\infty }p\left(x\right)dx=1 \\ Theintegralcanbesubdividedintermsoftwointegral, \\ {\int }_{-\infty }^{1}p\left(x\right)dx+{\int }_1^{\infty }p\left(x\right)dx=1 \\ Now,Substitutevalueofp\left(x\right) \\ {\int }_{-\infty }^1\left(0\right)dx+{\int }_1^{\infty }\left(\frac{c}{x^3}\right)dx=1 \\ {\left[\frac{c}{-2x^2}\right]}_1^{\infty }=1 \\ c=2 \end{gathered}$$

now, calculate probability of finding particle in interval 2mm$\le x\le 4mm$as follows,

$$\begin{gathered} probability(2mm\le x\le 4mm)={\int }_2^4\left(\frac{c}{x^3}\right)dx \\ ={\left[\frac{c}{-2x^2}\right]}_2^4 \\ =\left[\frac{c}{-2{\left(4\right)}^2}-\frac{c}{-2{\left(2\right)}^2}\right] \\ =\frac{c\left(3\right)}{32} \\ Now,substitute,2forcintheaboveresult \\ probability(2mm\le x\le 4mm)=\frac{\left(2\right)\left(3\right)}{32} \\ =\frac{3}{16} \\ Hence,probabilityoffindingparticleininterval2mm\le x\le 4mmis\frac{3}{16} \end{gathered}$$