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Q14Q

Expert-verifiedFound in: Page 1181

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Figure 38-24 shows an electron moving through several regions where uniform electric potentials V have been set up. Rank the three regions according to the de Broglie wavelength of the electron there, greatest first.**

The rank is ${\lambda}_{2}>{\lambda}_{1}>{\lambda}_{3}$.

**The de Broglie wavelength is given by,**

**${\mathit{\lambda}}{\mathbf{=}}\frac{\mathbf{h}}{\mathbf{p}}$**

**Here, h is the Planck's constant, ${\mathit{\lambda}}$ is wavelength, and p is the momentum. **

**The kinetic energy in terms of momentum is given by,**

**role="math" localid="1663142299185" ${\mathit{K}}{\mathbf{=}}\frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\phantom{\rule{0ex}{0ex}}{\mathit{p}}{\mathbf{=}}\sqrt{\mathbf{2}\mathbf{m}\mathbf{K}}$**

**Here, m is the mass.**

**Consider an electron, accelerated with a potential difference of V, the kinetic energy is given by,**

**${\mathit{K}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathit{e}}{\mathit{V}}$**

**The momentum is given by,**

**${\mathit{p}}{\mathbf{=}}\sqrt{\mathbf{2}\mathbf{m}\mathbf{e}\mathbf{V}}$**

**The expression for wavelength will be as follows:**

${\mathit{\lambda}}{\mathbf{=}}\frac{\mathbf{h}}{\sqrt{\mathbf{2}\mathbf{m}\mathbf{e}\mathbf{V}}}$

From the above de Broglie wavelength expression, the wavelength is inversely proportional to the $\sqrt{V}$, which means the wavelength decreases as the potential difference increases and vice-versa.

As given the potential difference,

${V}_{1}=-100\text{}V\phantom{\rule{0ex}{0ex}}{V}_{2}=-200\text{}V\phantom{\rule{0ex}{0ex}}{V}_{3}=+100\text{}V\phantom{\rule{0ex}{0ex}}$

So, the rank of de Broglie wavelength will be ${\lambda}_{2}>{\lambda}_{1}>{\lambda}_{3}$ .

Therefore, the rank is ${\lambda}_{2}>{\lambda}_{1}>{\lambda}_{3}$ .

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