Question

What is the wavelength of an object of a mass of $1.13 \times {10^4}g$ moving at a speed of $1.0 \times {10^3}cm/s$ ? (Planck’s constant $6.63 \times {10^{ - 27}}erg\sec $ ).
A. $5 \times {10^{ - 18}}cm$
B. $5 \times {10^{ - 30}}cm$
C. $5 \times {10^{ - 34}}cm$
D. $5 \times {10^{ - 38}}cm$

Answer 1

Hint:According to this question, we need to find the wavelength of an object whose mass and speed are given. So, here we will solve this question with the help of deBroglie wavelength as we have the planck’s constant, mass of the object and velocity of the object.

Complete step by step answer:
 De Broglie wavelength is a wavelength manifested in all the objects in quantum mechanics which determines the probability density of finding the object at a given point of the configuration space. The de Broglie wavelength of a particle is inversely proportional to its momentum. de Broglie equation states that matter can behave as waves much like light and radiation, which also behave as waves and particles.

The equation also explains that a beam of electrons can also be diffracted just like a beam of light. In essence, the de Broglie equation helps us to understand the idea of matter having a wavelength. 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.
Wavelength, $\lambda = \dfrac{h}{{mv}}$
where $h$= Planck's constant $m$= mass and $v$= velocity.
On putting the values, we get,
$\lambda = \dfrac{{6.626 \times {{10}^{ - 27}}}}{{1.3 \times {{10}^4} \times 1.0 \times {{10}^4}}}$
$\therefore \lambda = 5 \times {10^{ - 34}}cm$

Therefore, the correct answer is option C.

Note:De Broglie wavelength is a wavelength manifested in all the objects in quantum mechanics which determines the probability density of finding the object at a given point of the configuration space. De Broglie wavelength is only used when both mass of the object and the velocity of the object is given. We must remember the formula for de broglie wavelength $\lambda = \dfrac{h}{{mv}}$ and should provide the final answer after cancelling the common factors in numerator and denominator.