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Q13P

Expert-verifiedFound in: Page 1215

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**one-dimensional infinite well of length 200 pm contains an electron in its third excited state. We position an electron detector probe of width 2.00 pm so that it is centred on a point of maximum probability density. (a) What is the probability of detection by the probe? (b) If we insert the probe as described 1000 times, how many times should we expect the electron to materialize on the end of the probe (and thus be detected)?**

(a) The probability of detection is 0.020.

(b) 20 times should we expect the electron to materialize on the end of

the probe.

**As the gas decays and gives off electrons, the detector uses a magnet to trap them in a magnetic bottle. A radio antenna then picks up very weak signals emitted by the electrons, which can be used to map the electrons' precise activity over several milliseconds.**

Probability of defection in width centred on position is given by

$p\left(x\right)={\psi}_{n}^{2}\left(x\right)dx$

Probability density $\psi {}_{n}{}^{2}\left(x\right)$ for the trapped electron is

${\psi}_{n}^{2}\left(x\right)={A}^{2}si{n}^{2}\left(\frac{n\pi}{L}x\right),\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}[forn=1,2,3,...]$

Centred of the well, $x=\frac{L}{2}$

**$\begin{array}{l}=\frac{200pm}{2}\\ =100pm\end{array}$ **

** **

Probability of detection at x ,

**$p\left(x\right)={\psi}_{n}^{2}\left(x\right)dx\phantom{\rule{0ex}{0ex}}={\left[\sqrt{\frac{2}{L}\mathrm{sin}\left(\frac{n\mathrm{\pi}}{L}x\right)}\right]}^{2}dx\phantom{\rule{0ex}{0ex}}=\frac{2}{L}{\mathrm{sin}}^{2}\left(\frac{n\mathrm{\pi}}{L}x\right)dx$ **

** **

Here,

*n=3*

Length, $L=200pm$

Width of the probe, dx=2.00 pm

Probability

$\mathrm{p}\left(\mathrm{x}=\frac{\mathrm{L}}{2}\right)=\frac{2}{\mathrm{L}}{\mathrm{sin}}^{2}\left(\frac{\mathrm{n\pi}}{\mathrm{L}}\mathrm{x}\right)\mathrm{dx}\phantom{\rule{0ex}{0ex}}=\frac{2}{\mathrm{L}}{\mathrm{sin}}^{2}\left[\left(\frac{3\mathrm{\pi}}{\mathrm{L}}\right)\left(\frac{\mathrm{L}}{2}\right)\right]\mathrm{dx}\phantom{\rule{0ex}{0ex}}=\frac{2}{\mathrm{L}}{\mathrm{sin}}^{2}\left(\frac{3\mathrm{\pi}}{2}\right)\mathrm{dx}\phantom{\rule{0ex}{0ex}}=\frac{2}{\mathrm{L}}\mathrm{dx}\phantom{\rule{0ex}{0ex}}=\frac{2}{200\mathrm{pm}}\left(2.00\mathrm{pm}\right)\phantom{\rule{0ex}{0ex}}=0.020$

Hence, the probability of detection is 0.020.

Number of independent insertion,

$N=1000$

Number of times the electron to be detected, $n=Np$

$=\left(1000\right)\left(0.020\right)\phantom{\rule{0ex}{0ex}}=20$

Hence, 20 times should we expect the electron to materialize on the end of

the probe.

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