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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