Respuesta :
Answer:
[tex]\frac{1}{3}[/tex]
Step-by-step explanation:
From the Schrodinger equation, we have that the the wave function [tex]\psi[/tex] for a particle in 1-dimension box of length a, with no forces acting on it, is given by
[tex]\psi = \sqrt{\frac{2}{a} } \sin \left( \dfrac{n\pi x}{a} \right)[/tex]
where
- [tex]a[/tex] is the length of the box
- [tex]n[/tex] is the energy level
- [tex]x[/tex] is a variable depending on the size of the box
In this case, we have a state corresponding to [tex]n = 10[/tex]. Hence,
[tex]\psi = \sqrt{\frac{2}{a} } \sin \left( \dfrac{10\pi x}{a} \right)[/tex]
To find the probability that a particle will be found in a region between [tex]\frac{a}{3}[/tex] and [tex]\frac{2a}{3}[/tex] , we need to calculate the following integral
[tex]p = \int\limits^{\frac{a}{3}}_{\frac{2a}{3}} \psi \psi ^* \, dx[/tex]
where [tex]\psi^*[/tex] is the conjugate wave equation of [tex]\psi[/tex] . Thus,
[tex]\psi ^* = \sqrt{\frac{2}{a} } \sin \left( \dfrac{n\pi x}{a} \right)[/tex]
and we have
[tex]p = \int\limits^{\frac{a}{3}}_{\frac{2a}{3}} \sqrt{\frac{2}{a} } \sin \left( \dfrac{n\pi x}{a} \right) \cdot \sqrt{\frac{2}{a} } \sin \left( \dfrac{n\pi x}{a} \right) \, dx \\\phantom{p} = \frac{2}{a} \int\limits^{\frac{a}{3}}_{\frac{2a}{3}} \sin ^2\left( \dfrac{n\pi x}{a} \right) \, dx[/tex]
To solve the integral above, use the trigonometric identity
[tex]\sin ^2 x = \dfrac{1 - \cos 2x}{2}[/tex]
Therefore,
[tex]p = \frac{2}{a} \int\limits^{\frac{a}{3}}_{\frac{2a}{3}} \frac{1- \cos 2\left( \frac{n\pi x}{a} \right)}{2} \, dx\\\phantom{p} = \frac{2}{a} \int\limits^{\frac{a}{3}}_{\frac{2a}{3}} \frac{dx}{2} - \frac{2}{a} \int\limits^{\frac{a}{3}}_{\frac{2a}{3}} \frac{\cos 2\left( \frac{n\pi x}{a} \right)}{2} \, dx\\\phantom{p} = \frac{1}{a} x \Big|\limits^{\frac{a}{3}}_{\frac{2a}{3}} \\\phantom{a} = \frac{1}{a} \left(\frac{2a}{3} - \frac{a}{3} \right) \\\phantom{a} = \frac{1}{3}[/tex]
