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Q71CE

Expert-verifiedFound in: Page 1

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**(a) Find the wavelength of a proton whose kinetic energy is equal 10 its integral energy. **

**(b) ' The proton is usually regarded as being roughly of radius** ${\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{m}$**. Would this proton behave as a wave or as a particle?**

(a)** **The wavelength of a proton that has kinetic energy equal to its internal energy is $7.63\times {10}^{-16}\mathrm{m}$

(b) The moving proton would behave like as a particle in nature.

$\mathrm{Mass}\mathrm{of}\mathrm{proton},m=1.67\times {10}^{-27}\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Speed}\mathrm{of}\mathrm{light},c=3.0\times {10}^{8}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Plank}\text{'}\mathrm{s}\mathrm{constant},h=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}$

**The internal energy E**

** ${\mathit{E}}{\mathbf{=}}{\mathit{m}}{{\mathit{c}}}^{{\mathbf{2}}}$**

**The equation for the kinetic energy ( KE) **

**${\mathit{K}}{\mathit{E}}{\mathbf{=}}{\mathbf{(}}{{\mathit{\gamma}}}_{{\mathbf{u}}}{\mathbf{-}}{\mathbf{1}}{\mathbf{)}}{\mathit{m}}{{\mathit{c}}}^{{\mathbf{2}}}$ **

**Where, m**

** **

**De Broglie's wavelength**

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

**Where, p**

** **

**Relativistic effect in de Broglie's equation**

**${\mathit{\lambda}}{\mathbf{=}}\frac{\mathbf{h}}{{\mathbf{\gamma}}_{\mathbf{u}}\mathbf{m}\mathbf{v}}$ **

**Where, **** is the velocity at which the protons kinetic energy equals its internal energy, and ${\mathit{\gamma}}$**** is Lorentz factor.**

**Speed at which the internal energy become equal to the kinetic energy**

**${\mathit{m}}{{\mathit{c}}}^{{\mathbf{2}}}{\mathbf{=}}{\left({\gamma}_{u}-1\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{1}}{\mathbf{=}}{{\mathit{\gamma}}}_{{\mathbf{u}}}{\mathbf{-}}{\mathbf{1}}\phantom{\rule{0ex}{0ex}}{\mathbf{2}}{\mathbf{=}}{{\mathit{\gamma}}}_{{\mathbf{u}}}$ **

**Lorentz factor is given by a relation,**

** ${{\mathit{\gamma}}}_{{\mathbf{u}}}{\mathbf{=}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{2}}{\mathbf{=}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}\frac{{v}^{2}}{{c}^{2}}}}\phantom{\rule{0ex}{0ex}}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{1}}{\mathbf{-}}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{4}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{3}}{\mathbf{4}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathit{v}}{\mathbf{=}}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}{\mathit{c}}{\mathbf{}}$ **

(a)

Substitute the value of velocity in wavelength $\lambda $,

The wavelength of a proton that has kinetic energy equal to its internal energy is given by,

$\lambda =\frac{h}{\gamma m\left(\frac{\sqrt{3c}}{2}\right)}$

Substitute $6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{for}\mathrm{h},1.67\times {10}^{-27}\mathrm{kg}\mathrm{for}\mathrm{m}\mathrm{and}3.0\times {10}^{8}\mathrm{m}/\mathrm{s}$for *c* in the above equation to solve for $\lambda $

$\lambda =\frac{6.63\times {10}^{-34}J.s}{\left(2\right)(1.67\times {10}^{-27}\mathrm{kg})(\sqrt{\frac{3\times 3.0\times {10}^{8}\mathrm{m}/\mathrm{s}}{2}}}\phantom{\rule{0ex}{0ex}}=7.63\times {10}^{-16}\mathrm{m}$

Hence, the wavelength of a proton that has kinetic energy equal to its internal energy is $\lambda =7.63\times {10}^{-16}\mathrm{m}$

(b)

The wavelength of a proton that has kinetic energy equal to its internal energy is $\mathrm{\lambda}=7.63\times {10}^{-16}\mathrm{m}$

As the wavelength of the proton is smaller than the roughly size of the proton .

$\left({10}^{-15}\mathrm{m}\right)$So, the moving proton would behave like as a particle in nature.

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