一粒子在一维无限深方势阱中运动而处于基态。试求(1) 坐标$x$的平均值
A: $a$
B: $\dfrac{a}{2}$
C: $\dfrac{a}{4}$
D: $\dfrac{3a}{4}$
A: $a$
B: $\dfrac{a}{2}$
C: $\dfrac{a}{4}$
D: $\dfrac{3a}{4}$
举一反三
- （接上题）（2）设经分界面反射的波的振幅和入射波的振幅相等，则反射波的波函数是 A: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{2} \right),0\le x\le\dfrac{3\lambda}{4}$ B: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$ C: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{4} \right),0\le x\le\dfrac{3\lambda}{4}$ D: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$
- （2）设经分界面反射的波的振幅和入射波的振幅相等，则反射波的波函数是 A: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{2} \right),0\le x\le\dfrac{3\lambda}{4}$ B: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$ C: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{4} \right),0\le x\le\dfrac{3\lambda}{4}$ D: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$
- 设粒子在一维无限深方势阱中运动并处于基态.求在势阱中距势阱内壁1/4宽度以内发现粒子的概率。
- \(t\sin(2t)\)的拉普拉斯变换为 A: \(\dfrac{2}{s^2+4}\) B: \(\dfrac{s}{s^2+4}\) C: \(\dfrac{4}{(s^2+4)^2}\) D: \(\dfrac{4s}{(s^2+4)^2}\)
- 函数 $f(x)=2x^2-x+1$ 在区间 $[-1,3]$ 上满足拉格朗日定理的 $\xi$ 等于( ). A: $-\dfrac{3}{4}$ B: $0$ C: $\dfrac{3}{4}$ D: $1$