Question

Find the probability that a particle in a box L wide can be found between x = 0 and x = 1/3 when he particle is in the 3rd excited state.

Answer 1

Answer:

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Answer 2

Answer:

In a one dimensional box of length, L a particle is there in the 3rd excited state​ the probability of finding the particle in between 0 and L/3 is 0.29

Explanation:

• The wavefunction of a particle gives the quantum state of that particle, for a particle in a one-dimensional box of length L the wavefunction is

         $\ \psi \left(x\right)=\sqrt{2/L}\ sin \left(\frac{n\pi x}{L}\right)$       $n$- state of the particle

• The probability of finding the particle within the limit a to b is

         $\int _a^b\psi \left(x\right)^{\ast }\psi \left(x\right)dx=\frac{2}{L}\int _a^b\sin ^2\frac{n\pi x}{L}dx$

• we have a trigonometric formula

          $sin^{2} \alpha =(cos2\alpha -1)/2$

From the question, we have given

the state of the particle $n=4$ (third excited state)

the probability of particle within the limit $0$ to $L/3$  in the third excited state is

$\int _0^\frac{L}{3} \psi \left(x\right)^{\ast }\psi \left(x\right)dx=\frac{2}{L}\int _0^\frac{L}{3} \sin ^2\frac{4\pi x}{L}dx$

using the trigonometric formula,

⇒$\frac{2}{2L}\int _0^\frac{L}{3} (\cos\frac{8\pi x}{L}-1)dx$

split the integral

$\frac{1}{L}\int _0^{\frac{L}{3}}\cos \frac{8\pi x}{L}dx-\frac{1}{L}\int _0^{\frac{L}{3}}dx$

$\frac{1}{L}\frac{L}{8\pi } \left[sin(8\pi x/L)\right]_0^{\frac{L}{3}}-\frac{1}{L}\left[x\right]_0^{\frac{L}{3}}$

now substitute the limits

$\frac{1}{8\pi } \left[sin(8\pi/3-sin0 )\right]}-\frac{1}{L}\left[L/3-0\right]}=0.29$

therefore the probability is $0.29$