To determine the wavelength associated with the given objects, we can use the de Broglie wavelength formula:
λ = h / (mv)
Where λ is the wavelength, h is the Planck's constant (6.626 × 10^(-34) J·s), m is the mass of the object, and v is its velocity.
For the bullet: m = 60 gm = 0.06 kg v = 250 m/s
Plugging these values into the equation, we get: λ = (6.626 × 10^(-34) J·s) / ((0.06 kg) × (250 m/s))
Calculating this expression gives us the de Broglie wavelength associated with the bullet.
For the electron, the mass can vary depending on whether it is a free electron or part of an atom. For a free electron, the mass is approximately 9.11 × 10^(-31) kg.
Let's assume it's a free electron with: m = 9.11 × 10^(-31) kg v = 250 m/s
Using the same formula, we can calculate the de Broglie wavelength associated with the electron.
Regarding the second part of your question, the wave nature of bullets is not typically revealed through diffraction effects due to their relatively large mass and momentum compared to fundamental particles like electrons. Diffraction occurs when a wave encounters an obstacle or passes through a narrow opening, causing it to spread out and interfere with itself. This phenomenon is more noticeable in particles with smaller mass, such as electrons or photons, which have wave-like properties. Bullets, on the other hand, have much larger mass and momentum, making their associated wavelengths extremely small and difficult to observe diffraction effects in everyday situations.