Question

What is the de Broglie wavelength of?
(A) A bullet of mass $0.040$ kg travelling at the speed of $1.0{\text{ km}}{{\text{s}}^{ - 1}}$
(B) A ball of mass $0.060{\text{ kg}}$ moving at a speed of $1.0{\text{ m}}{{\text{s}}^{ - 1}}$
(C) A dust particle of mass $1 \times {10^9}{\text{ kg}}$ drifting with a speed of $2.2{\text{ m}}{{\text{s}}^{ - 1}}$ .

Answer 1

Hint : De- Broglie equation holds account for the wave nature of the particle. So basically De-Broglie wavelength describes the wave-like nature of the smaller particle electron. And the wavelength calculated using the De—Broglie equation is known as the De-Broglie wavelength.

Complete Step By Step Answer:
As we know that matter can exist in both nature i.e. can exist in the particle as well as in the wave-like form.
So, as Einstein's equation describes the particle nature of matter, in the same way, De-Broglie wavelength describes the wave-like nature of the particle or the electron.
Now, we will have a look at the de-Broglie wavelength equation:
$\Rightarrow$ $\lambda = \dfrac{h}{p}$
Here, $\lambda$ stands for the wavelength of the particle
$h$ Is the Planck’s constant
$p$ Is the momentum of the particle?
Since, momentum of the particle is:
$\Rightarrow$ $p = mv$
m is the mass of the particle
v is the velocity of the particle
So, De-Broglie wavelength can also be written as:
$\Rightarrow$ $\lambda = \dfrac{h}{{mv}}$
Now, putting our values in the De-Broglie wavelength equation, we will get our answer:
$m = 0.040kg$ $v = 1.0km{s^{ - 1}}$
So, putting the values in equation:
$\lambda = \dfrac{h}{{mv}}$
$\lambda = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{0.04 \times 1000}}$
$= 1.655 \times {10^{ - 35}}m$
$\lambda = \dfrac{h}{{mv}}$
So, putting the values in equation:
$\lambda = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{0.06 \times 1}}$
$= 1.1 \times {10^{ - 32}}m$
$\lambda = \dfrac{h}{{mv}}$
So, putting the values in equation:
$\lambda = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{1 \times {{10}^{ - 9}} \times 2.2}}$
$= 3 \times {10^{ - 23}}m$
So, we obtained the De-Broglie wavelength.

Note :
Important point to note here is that in the first option we have multiplied the denominator with $1000$ , this is because the velocity is given in $1.0{\text{ km}}{{\text{s}}^{ - 1}}$ i.e. in kilometre per second so we will convert it in the standard units i.e. meter per second . So to convert kilometers per second into meters per second we will multiply it by $1000$ .