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Multiple choice
For an electron in the 1D infinite square well ground state (
n
=
1
n=1
n
=
1
,
L
=
1
L=1
L
=
1
), the probability density is
∣
ψ
1
(
x
)
∣
2
=
2
sin
2
(
π
x
)
|\psi_1(x)|^2 = 2 \sin^2(\pi x)
∣
ψ
1
(
x
)
∣
2
=
2
sin
2
(
π
x
)
. What is the probability of finding the particle in the left half of the well, i.e. the integral from
x
=
0
x=0
x
=
0
to
x
=
0.5
x=0.5
x
=
0.5
?
🔬 try it before you answer
1
/
2
=
0.5
1/2 = 0.5
1/2
=
0.5
2
/
π
≈
0.637
2/\pi \approx 0.637
2/
π
≈
0.637
1
/
π
≈
0.318
1/\pi \approx 0.318
1/
π
≈
0.318
1
/
4
=
0.25
1/4 = 0.25
1/4
=
0.25
Check answer