Question

A 200g cricket ball is thrown with a speed of $3.0 \times {10^3}cm/s$. What will be its de Broglie’s wavelength?
$(h = 6.6 \times {10^{ - 27}}gc{m^2}{\sec ^{ - 1}})$
a.) $1.1 \times {10^{ - 32}}cm$
b.) $2.2 \times {10^{ - 32}}cm$
c.) $0.55 \times {10^{ - 32}}cm$
d.) $11.0 \times {10^{ - 32}}cm$

Answer 1

Hint: This question can be solved by using the de Broglie equation, that is $\lambda = \dfrac{h}{{mv}}$ , where h is the Planck's constant, $\lambda $ is the de Broglie wavelength and $m$ is the mass of the particle moving at a velocity $v$ .

Complete step-by-step answer:
Given in the question that the speed of the ball i.e. $v$ is $3.0 \times {10^3}cm/s$. Also given the mass i.e. $m$ is 200g and the value of planck's constant $h = 6.6 \times {10^{ - 27}}gc{m^2}{\sec ^{ - 1}}$.
De Broglie’s equation relates a moving particle’s wavelength to its momentum. The equation, therefore becomes, $\lambda = \dfrac{h}{p}$ , where h is the Planck's constant, and p is the momentum of the particle. The momentum of any object or particle is given by $p = mv$ . Therefore, the de Broglie equation becomes $\lambda = \dfrac{h}{{mv}}$.
Substituting the values of planck's constant, mass and velocity in the de Broglie equation, i.e. $\lambda = \dfrac{h}{{mv}}$, we get
$ \Rightarrow \lambda = \dfrac{{6.6 \times {{10}^{ - 27}}gc{m^2}{{\sec }^{ - 1}}}}{{200g \times 3 \times {{10}^3}cm/s}}$
Further simplifying,
$ \Rightarrow \lambda = \dfrac{{2.2 \times {{10}^{ - 30}}cm}}{{200}}$
$ \Rightarrow \lambda = 1.1 \times {10^{ - 32}}cm$
Hence, de Broglie’s wavelength is calculated to be $1.1 \times {10^{ - 32}}cm$.
Therefore, option A is correct.

Note- de Broglie’s equation describes the wave property of matter, especially the wave nature of electrons. De Broglie suggested that a particle can exhibit properties of a wave. The matter has a dual nature. This is also known as de Broglie hypothesis. De Broglie wavelength is a wavelength exhibited in all the objects in quantum physics which establishes the probability density of discovering the object at a given point of the arrangement space. The de Broglie wavelength of a particle is inversely proportional to its momentum.