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A 10 kg ball is released from the top of a 235 m tall building. What is its De Broglie wavelength just before hitting the Earth?

Answer :

The De Broglie wavelength of the 10 kg ball just before hitting the earth is approximately 6.63 x 10^(-37) meters.

The De Broglie wavelength is calculated using the equation:

λ = h / p

Where:

λ is the De Broglie wavelength,

h is Planck's constant (approximately 6.626 x 10^(-34) J·s),

p is the momentum of the object.

To calculate the momentum, we can use the equation:

p = m * v

Where:

m is the mass of the object,

v is the velocity of the object.

Mass of the ball (m) = 10 kg

Height of the building (h) = 235 m

To find the velocity of the ball just before hitting the earth, we can use the equation for gravitational potential energy:

m * g * h = (1/2) * m * v^2

Where:

g is the acceleration due to gravity (approximately 9.8 m/s^2).

Simplifying the equation:

v^2 = 2 * g * h

Substituting the given values:

v^2 = 2 * 9.8 m/s^2 * 235 m

v^2 ≈ 4,607 m^2/s^2

Taking the square root:

v ≈ √(4,607 m^2/s^2)

v ≈ 67.87 m/s

Now we can calculate the momentum:

p = m * v

p = 10 kg * 67.87 m/s

p ≈ 678.7 kg·m/s

Finally, we can calculate the De Broglie wavelength:

λ = h / p

λ = 6.626 x 10^(-34) J·s / 678.7 kg·m/s

λ ≈ 9.75 x 10^(-37) meters

Therefore, the De Broglie wavelength of the 10 kg ball just before hitting the earth is approximately 6.63 x 10^(-37) meters.

The De Broglie wavelength of the 10 kg ball just before hitting the earth, calculated using the given mass and the height of the building, is approximately 6.63 x 10^(-37) meters. This calculation is based on the principles of quantum mechanics and the equations for momentum and De Broglie wavelength.

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