Chapter 28, Problem 2P

To determine

Lengthof hole.

Answer to Problem 2P

The answer is $\text{1}\text{.2}\text{×}{\text{10}}^{\text{28}}\text{m}$

Explanation of Solution

Given info: Mass of bullet, $\text{m = 3}\text{.0}\text{g}\text{=}\text{3}{\text{.0×10}}^{\text{-3}}\text{kg}$

Speed of bullet, $\text{v = 220}\text{m/s}$

Diameter of hole, $\text{D = 3}\text{×}{\text{10}}^{\text{-3}}\text{m}$

Radius is $\frac{\text{D}}{\text{2}}\text{=}\frac{\text{1}\text{.0}\text{cm}}{\text{2}}\text{=0}\text{.50}\text{cm}\text{=}\text{0}{\text{.5×10}}^{\text{-2}}\text{m}$

Formula used:

The wavelength is $\text{λ=}\frac{\text{h}}{\text{p}}\text{=}\frac{\text{h}}{\text{mv}}$

  $\text{λ}$ is wavelength

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

  $p$ is photon’s momentum

  $v$ is velocity.

  $m$ is particle’s mass

Calculation:

First calculate the wavelength of bullet by using the formula $\text{λ=}\frac{\text{h}}{\text{mv}}$

  $\text{=}\frac{\left(\text{6}{\text{.63×10}}^{\text{-34}}\text{J×s}\right)}{{\left[\left(\text{3}{\text{.0×10}}^{\text{-3}}\text{kg}\right)\left(\text{220}\text{m/s}\right)\right]}^{\frac{\text{1}}{\text{2}}}}\text{=1}{\text{.0×10}}^{\text{-33}}\text{m}$

Understand that the half-angle for the central circle of the diffraction pattern is given as $\text{sin}\text{θ=}\frac{\text{1}\text{.22}\text{λ}}{\text{D}}$

For small angles $\text{sin}\text{θ}\approx \text{tanθ}$

So, $\text{r}\text{=}\text{L}\text{tan}\text{θ}\approx \text{L}\text{sin }\text{θ}=\frac{\text{1}\text{.22}\text{L}\text{λ}}{\text{D}}\mathrm{...}\left(1\right)$

Substitute the all the values in equation $\left(1\right)$

  $\text{0}{\text{.50×10}}^{\text{-2}}\text{m = }\frac{\text{1}\text{.22}\text{L}\left(\text{1}\text{.0}{\text{×10}}^{\text{-33}}\text{m}\right)}{\left(\text{3}{\text{.0×10}}^{\text{-3}}\text{m}\right)}$

  $\text{L = 1}\text{.2}\text{×}{\text{10}}^{\text{28}}\text{m}$

Conclusion:

The required value is $\text{1}\text{.2}\text{×}{\text{10}}^{\text{28}}\text{m}$ .