91影视

The quantitative value of the wavelength$k=20{\text{nm}}^{鈭�1}$

Formula used:

$\mathit{k}\mathbf{=}\frac{\mathbf{2}\mathbf{蟺}}{\mathbf{位}}$,

Where$\mathit{k}$ is quantitative value of the wavelength, $\mathit{位}$is the wavelength.

(a)

The quantitative value of the wavelength will correspond to the high contribution coming from the central value at$k=20{\text{nm}}^{鈭�1}$.

Hence, the wavelength will be given as follows.

Substitute the value,

$位=\frac{2蟺}{20脳{10}^9{\text{m}}^{鈭�1}}=0.314\text{nm}$

For the position uncertainty, use uncertainty relation for the Fourier transformation (with $k$instead of$p$}

$\begin{array}{c}螖x螖k=\frac{1}{2}\\ 螖x=\frac{1}{2螖k}螖x\\ =\frac{1}{2脳I_{{\text{nm}}^{鈭�1}}}螖x\\ =0.5\text{nm}\end{array}$

Where the value of$螖k$ could be read from the graph, corresponding to the width of the curve at$\text{1/e}$ of the peak value, it tunes out to be approximately.

(b)

Unfortunately, the particle is represented as a wave, hence the wave loses its particle nature in the first place there. Nevertheless, if it is transformed to the k-space, it is only the wave number that is not sufficient to give the approximate position of the particle. At most it is possible to find the certainty in position as in part a, however, the position itself is inconclusive.