A Bullet's Wavelength

Q. What is the de Broglie wavelength of a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, in m? expects 1.66 × 10⁻³⁵ m

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Q. What is the de Broglie wavelength of a bullet of mass 0.040 kg $m = 0.040$ kg travelling at the speed of 1.0 km/s $v = 1.0$ km/s, in m?

The playground

Inputs

mass

m

0.04 kg

speed

v

1 km/s

Outputs

wavelength

λ

The solution

Solved with MatterWaves

Given

$m = 0.040$ kg
$v = 1.0$ km/s

Find $λ$ .

Working

$v = 1.0$ km/s $= 1000$ m/s
de broglie relation(particle) $λ = \frac{h}{m \cdot v}$
$λ = \frac{6.62607015 \times 10^{−34}}{0.04 \cdot 1000}$
$λ = 1.657 \times 10^{−35}$ m
$λ \approx 1.66 \times 10^{−35}$ mto 3 significant figures

Conclusion

A bullet's momentum is enormous next to Planck's constant, so its wavelength is far too small to show any wave behaviour.

The Ruby

1require_relative "../physics"23question :a_bullet_s_wavelength do4  kind :numerical5  describes "%{wavelength} of a bullet of mass %{mass} travelling at the speed of %{speed}, in m?"6  given mass: "0.040 kg"7  given speed: "1.0 km/s"8  asks wavelength: "What is the de Broglie wavelength"9end

1require_relative "../questions/a_bullet_s_wavelength"2require_relative "../science/dual_nature_of_radiation_and_matter/matter_waves"34solve :a_bullet_s_wavelength, using: MatterWaves do5  objects :particle67  take mass: particle.mass8  take speed: particle.speed910  wavelength = particle.de_broglie_wavelength11  answer wavelength:, in: "m", figures: 31213  conclusion "A bullet's momentum is enormous next to Planck's constant, so its wavelength is far too small to " \14             "show any wave behaviour."15end

1require_relative "../solutions/a_bullet_s_wavelength"23play_with :a_bullet_s_wavelength do4  title "A Bullet's Wavelength"5end

The laws

quantity	written	units	within	from
plancks constant	$h$	J s	—	MatterWaves
mass	$m$	kg	0 … ∞	MatterWaves
speed	$v$	m/s	—	MatterWaves
de broglie wavelength	$λ$	m	—	MatterWaves
de broglie wavelength of ground	$λ_{g}$	m	—
mass of ground	$m_{g}$	kg	0 … ∞
speed of ground	$v_{g}$	m/s	—
de broglie wavelength of particle	$λ$	m	—
mass of particle	$m$	kg	0 … ∞
speed of particle	$v$	m/s	—

λ_{p} = \frac{h}{m_{p} \cdot v_{p}}

de broglie relation

λ_{g} = \frac{h}{m_{g} \cdot v_{g}}

de broglie relation(ground)

λ = \frac{h}{m \cdot v}

de broglie relation(particle)