Question

what will be the probability of finding a particle between 0 to 0.5l in one dimension box of length l in the second excited state​

Answer 1

Answer:

In a one dimensional box of length, L a particle is there in the second excited state​ the probability of finding the particle in between 0 and L/2 is $0.5$

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 $n=3$

the probability of particle within the limit $0$ to $L/2$  in the second excited state is

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

using the trigonometric formula,

⇒$\frac{2}{2L}\int _0^\frac{L}{2} (\cos\frac{6\pi x}{L}-1)dx$

$\frac{1}{L}\int _0^\frac{L}{2} (\cos\frac{6\pi x}{L}-1)dx$

split the integral

$\frac{1}{L}\int _0^{\frac{L}{2}}\cos \frac{6\pi x}{L}dx-\frac{1}{L}\int _0^{\frac{L}{2}}dx$

$\frac{1}{L}\frac{L}{6\pi } \left[sin(6\pi x/L)\right]_0^{\frac{L}{2}}-\frac{1}{L}\left[x\right]_0^{\frac{L}{2}}$

now substitute the limits

$\frac{1}{6\pi } \left[sin(3\pi-sin0 )\right]}-\frac{1}{L}\left[L/2-0\right]}=1/2$

therefore the probability is $0.5$