Question

What is the de Broglie wavelength of a ball with mass of $ 0.20kg $ when it strikes the ground after it has been dropped from a building that’s $ 50m $ tall?

Answer 1

Hint: From the given height from which the ball was dropped and acceleration due to gravity, the velocity can be calculated. By substituting the values of Planck’s constant, mass of a ball, and velocity in the formula of de Broglie wavelength gives the value of wavelength in metres.
 $ \lambda = \dfrac{h}{{m\nu }} $
 $ \lambda $ is De Broglie wavelength
 $ h $ is Planck’s constant
 $ m $ is mass of a ball
 $ \nu $ is velocity having to be determined.

Complete Step By Step Answer:
Given that the ball has been dropped from a building that’s $ 50m $ tall. let the height be $ x $
The acceleration due to gravity is a constant value and equal to $ 9.8m{s^{ - 2}} $
By substituting these both values in the velocity which is given as
 $ \nu = \sqrt {2gx} $
The velocity can be written as $ \nu = \sqrt {2 \times 9.8 \times 50} = \sqrt {9860} {m^2}{s^{ - 2}} $
Further simplifying, $ \nu = 31.3m{s^{ - 1}} $
Given that the mass of a ball is $ 0.20kg $
Planck’s constant is a constant value which was given as $ 6.626 \times {10^{ - 34}}kg.{m^2}.{s^{ - 1}} $
Substitute the values of mass, Planck’s constant and the determined velocity in the de Broglie wavelength
 $ \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}kg.{m^2}.{s^{ - 1}}}}{{0.20kg \times 31.3m{s^{ - 1}}}} $
Further simplifying the above equation,
 $ \lambda = 1.06 \times {10^{ - 34}}m $
Thus, the de Broglie wavelength of a ball with mass of $ 0.20kg $ when it strikes the ground after it has been dropped from a building that’s $ 50m $ tall is $ 1.06 \times {10^{ - 34}}m $ .

Note:
While calculating the velocity, the height of a building must be in metres only, as the acceleration due to gravity is a constant value in metres per second square. The Planck’s constant value must be taken in the M.K.S. system only, as the mass is given in kilograms. Thus, the units of the quantities must be taken correctly.