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Question: A 200g cricket ball is thrown with a speed of \(3.0 \times {10^3}cm/s\). What will be its de Broglie...

A 200g cricket ball is thrown with a speed of 3.0×103cm/s3.0 \times {10^3}cm/s. What will be its de Broglie’s wavelength?
(h=6.6×10−27gcm2sec⁡−1)(h = 6.6 \times {10^{ - 27}}gc{m^2}{\sec ^{ - 1}})
a.) 1.1×10−32cm1.1 \times {10^{ - 32}}cm
b.) 2.2×10−32cm2.2 \times {10^{ - 32}}cm
c.) 0.55×10−32cm0.55 \times {10^{ - 32}}cm
d.) 11.0×10−32cm11.0 \times {10^{ - 32}}cm

Explanation

Solution

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

Complete step-by-step answer:
Given in the question that the speed of the ball i.e. vv is 3.0×103cm/s3.0 \times {10^3}cm/s. Also given the mass i.e. mm is 200g and the value of planck's constant h=6.6×10−27gcm2sec⁡−1h = 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, λ=hp\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=mvp = mv . Therefore, the de Broglie equation becomes λ=hmv\lambda = \dfrac{h}{{mv}}.
Substituting the values of planck's constant, mass and velocity in the de Broglie equation, i.e. λ=hmv\lambda = \dfrac{h}{{mv}}, we get
⇒λ=6.6×10−27gcm2sec⁡−1200g×3×103cm/s\Rightarrow \lambda = \dfrac{{6.6 \times {{10}^{ - 27}}gc{m^2}{{\sec }^{ - 1}}}}{{200g \times 3 \times {{10}^3}cm/s}}
Further simplifying,
⇒λ=2.2×10−30cm200\Rightarrow \lambda = \dfrac{{2.2 \times {{10}^{ - 30}}cm}}{{200}}
⇒λ=1.1×10−32cm\Rightarrow \lambda = 1.1 \times {10^{ - 32}}cm
Hence, de Broglie’s wavelength is calculated to be 1.1×10−32cm1.1 \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.