Answer on Question #85010, Physics / Quantum Mechanics
Question:
The wavefunction for a particle is defined by:
ψ ( x ) = { N cos ( 2 π x / L ) for − L / 4 ≤ x ≤ L / 4 \psi (x) = \left\{N \cos \left(2 \pi x / L\right) \text{ for } - L / 4 \leq x \leq L / 4 \right. ψ ( x ) = { N cos ( 2 π x / L ) for − L /4 ≤ x ≤ L /4
0 otherwise
Determine
i) the normalization constant N N N , and
ii) the probability that the particle will be found between x = 0 x = 0 x = 0 and x = L / 8 x = L / 8 x = L /8 . (5+5)
Solution:
By entering t = 2 x + 0.5 L t = 2x + 0.5L t = 2 x + 0.5 L , we get its range [ 0 ; L ] [0;L] [ 0 ; L ] and the wavefunction ψ ( t ) = N sin ( π t / L ) \psi (t) = N\sin (\pi t / L) ψ ( t ) = N sin ( π t / L ) . In this case
N = 2 L and the probability N = \sqrt{\frac{2}{L}} \quad \text{and} \quad \text{the probability} N = L 2 and the probability p = 0.75 L − 0.5 L L − sin 2 π ⋅ 0.75 − sin 2 π ⋅ 0.5 6.28 = 0.25 + 0.16 = 0.41 p = \frac{0.75L - 0.5L}{L} - \frac{\sin 2\pi \cdot 0.75 - \sin 2\pi \cdot 0.5}{6.28} = 0.25 + 0.16 = 0.41 p = L 0.75 L − 0.5 L − 6.28 sin 2 π ⋅ 0.75 − sin 2 π ⋅ 0.5 = 0.25 + 0.16 = 0.41
The answer:
The normalization constant N = 2 L N = \sqrt{\frac{2}{L}} N = L 2
The probability p = 0.41 p = 0.41 p = 0.41
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