Chapter 27.2, Problem 30SSC

To determine

To Calculate: The de Broglie wavelength of alpha particle.

Answer to Problem 30SSC

  $\text{8}\text{.34}\stackrel{\text{o}}{\text{A}}$

Explanation of Solution

Given data:

Mass of alpha particle, $m=6.6\times {10}^{-27}\text{kg}$

Speed of alpha particle, $v=120\text{m}\cdot {\text{s}}^{-1}$

Formula used:

de Broglie wavelength $(\lambda )$ of the alpha particle can be obtained as:

  $\lambda =\frac{h}{mv}$

Where,

  $\lambda =$ de Broglie wavelength

  $h=$ Planck’s constant

  $m=$ mass of alpha particle

  $v=$ Velocity of alpha particle

Calculation:

Substitute the known values and solve:

  $\begin{array}{c}\lambda =\frac{h}{mv}\\ \text{=}\frac{6.6\times {10}^{-34}}{6.6\times {10}^{-27}\times 120}\\ =8.33\times {10}^{-10}\text{m }\\ \text{=}\text{8}\text{.34}\stackrel{\text{o}}{\text{A}}\end{array}$

Conclusion:

Therefore, de Broglie wavelength of alpha particle is $\text{8}\text{.34}\stackrel{\text{o}}{\text{A}}$ .