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What is the de-Broglie wavelength approximately of a tennis ball of mass 60 g moving with a velocity of 10 meters per second? (Planck’s constant, h = $6.63 \times {10^{ - 34}}Js$)
(A) ${10^{ - 25}}$ metres
(B) ${10^{ - 33}}$ metres
(C) ${10^{ - 31}}$ metres
(D) ${10^{ - 16}}$ metres

Last updated date: 20th Jun 2024
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Hint:The de Broglie wavelength of a particle indicates the length scale at which wave-like properties are important for that particle. Therefore, if we look at every moving particle whether it is microscopic or macroscopic it will have a wavelength. In cases of macroscopic objects, the wave nature of matter can be detected or it is visible.

Complete step-by-step solution:The concept that matter behaves like a wave was proposed by Louis de Broglie in 1924. It is also referred to as the de Broglie hypothesis. Matter waves are referred to as de Broglie waves. De Broglie equation states that a matter can act as waves much like light and radiation which also behave as waves and particles. The equation further explains that a beam of electrons can also be diffracted just like a beam of light.
De Broglie wavelength is given by the formula; $\lambda = \dfrac{h}{{mv}}$
From the question the data filtered out is: $mass(m) = 0.06kg$,$velocity(v) = 10m/s$ and
h = $6.63 \times {10^{ - 34}}Js$. Substituting the values of mass, velocity and Planck’s constant in the formula for de-Broglie wavelength we get:
$\lambda = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{0.06 \times 10}} = 1.1 \times {10^{ - 33}}m$
Hence the approx. value of de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 meters per second is $\lambda = {10^{ - 33}}m$.

Hence the correct option is (B).

Note:Electron has the least mass, so its wavelength is maximum. The significance of de Broglie relation is that it is more useful to microscopic, fundamental particles like electrons.