Compare the de broglie wavelength of an electron moving at 1.30×107 miles per hour (5.81×106 m/s) to that of a bullet moving at 700 miles per hour (313 m/s) and a golf ball with a speed of 70.0 miles per hour (31.3 m/s).which region of the electromagnetic spectrum are each of these wavelengths near? a. ultraviolet 10-8 to 10-7 metersb. x-ray 10-11 to 10-8 metersc. gamma ray 10-16 to 10-11 metersd. smaller than 10-20 meters. cannot detect wave-like properties. only particle-like behavior will be observable. particle mass (kg) velocity (m/s) wavelength (m) regionelectron 9.11×10-31 5.81×106 1.90×10-3 313 ball 4.50×10-2 31.3

electron  λ = 12.5 nm , bullet  λ = 1.11 10⁻³³ m  and golf ball  λ = 4.7 10⁻³⁴ m

Explanation:

The Broglie wave duality principle states that all matter has wave and particle properties, it is expressed by the equation

      p = h / λ

Where lam is called broglie wavelength

Let's use the definition of momentum

       p = mv

Let's calculate the wavelengths

-Electron

     mv = h /λ

     λ = h / mv

     λ = 6.63 10⁻³⁴ / (9.1 10⁻³¹ 5.81 10⁶)  

     λ = 1.25 10⁻¹⁰ m

     λ = 12.5 nm

This is the X-ray region

-bullet

     λ = 6.63 10⁻³⁴ / (1.90 10⁻³  313)

     λ = 1.11 10⁻³³ m

It is too small, only particle characteristics are observed

-Golf ball

     λ = 6.63 10⁻³⁴ / (4.50 10⁻² 31.3)

    λ = 4.7 10⁻³⁴ m

Too small, only particle characteristics are visible

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s