Question

Calculate the probability of finding the particle
between x= 0 to x=L/100, in an infinite potential
well of 10 Angstrom width. Given that the particle is in 100 th excited state.​

Answer 1

Answer:

Explanation:

Calculate the probability of finding the particle

between x= 0 to x=L/100, in an infinite potential

well of 10 Angstrom width. Given that the particle is in 100 th excited state.​

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Answer 2

Answer:

In an infinite potential  well of 10 Angstrom width, a particle is in 100th excited state. The probability of finding the particle in between $x=0$ and $x=L/100$ is $0.09$

Explanation:

• The wavefunction of a particle gives the quantum state of that particle, for a particle in an infinite square well potential 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 $n=101$ (100th excited state)

the probability of particle within the limit $0$ to $L/100$  in the 100th excited state is

$\int _0^\frac{L}{100} \psi \left(x\right)^{\ast }\psi \left(x\right)dx=\frac{2}{L}\int _0^\frac{L}{100} \sin ^2\frac{101\pi x}{L}dx$

using the trigonometric formula,

⇒$\frac{2}{2L}\int _0^\frac{L}{100} (\cos\frac{202\pi x}{L}-1)dx$

split the integral

$\frac{1}{L}\int _0^{\frac{L}{100}}\cos \frac{202\pi x}{L}dx-\frac{1}{L}\int _0^{\frac{L}{100}}dx$

$\frac{1}{L}\frac{L}{202\pi } \left[sin(202\pi x/L)\right]_0^{\frac{L}{100}}-\frac{1}{L}\left[x\right]_0^{\frac{L}{100}}$

now substitute the limits

$\frac{1}{202\pi } \left[sin(202\pi/100-sin0 )\right]}-\frac{1}{L}\left[L/100-0\right]}=0.09$

therefore the probability is $0.09$