Question

9.) If a bullet of mass 90 grains is fired from a gun at 300 ft/s, what would the value of the physical used has a constant need to be in order to see quantum effects, presuming that instrumentation resolution of 1 picometer. (1 grain = 0.0647 grams, 1 foot = 0.3048 m) 10.)

Answer 1

According to de Broglie's postulate all particles have a wave like properties quantum mechanically. As well as all electromagnetic radiation has a particle like properties on the other hand.

Based on this wave particle duality principle one can calculate the wavelength of a moving particle or object.
In the above question due to such quantum effect the bullet carries a wave like properties after it is fired.

The given below calculation provides the measurement of the wavelength :

De Broglie's equation is

where p is the linear momentum of the moving object, h is the planck's constant

According to de Broglie's postulate all particles have a wave like properties quantum mechanically. As well as all electromagnetic radiation has a particle like properties on the other hand. Based on this wave particle duality principle one can calculate the wavelength of a moving particle or object. In the above question due to such quantum effect the bullet carries a wave like properties after it is fired. The given below calculation provides the measurement of the wavelength: De Broglie's equation is lambda = h/p where p is the linear momentum of the moving object, h is the Planck's constant h = 6.6260700410^-34 m^2 kg/s and lambda is the wavelength of the particle wave Now for the given information's lambda_bullet = 6.626 times 10^34/p bullet here_P bullet = m_bullet v_bullet as mass of the bullet is given 90 grains i. e 90 x 0.0647 = 5.823 gram or 0.005823 kg. and the velocity of the bullet is 300 ft/s i. e.

and $\lambda$ is the wavelength of the particle wave


Now for the given informations

$\lambda_{bullet}= \frac{6.626\times 10^{-34}}{p_{bullet}}$

here


as mass of th bullet is given 90 grains i.e 90 x 0.0647 = 5.823 gram or 0.005823 kg. and the velocity of the bullet is 300ft/s i.e. 300 x 0.3048 = 91.44 m/s

therefore ${p_{bullet}}=m_{bullet} v_{bullet} = (5.823\times 10^{-3}\times 91.44) kgm/s =0.5324 kgm/s$


Now,
$\lambda_{bullet}= \frac{6.626\times 10^{-34}}{0.5324}=1.244\times 10^{-33}m$

The wavelentgh is too short to measure and that is why we set as much as possible low resolution instrumentation.
In the qustion it is provided that the optimal resolution of all instrument is 1 picometer or $10^{-12}m$


Therefore the physical constant we have to consider, let k, will be  

$k(physical\;constant)= \frac{order\; of\;\lambda}{10^{-12}}=\frac{10^{-33}}{10^{-12}} = 10^{-21}$

where k is just a dimensionless number,
we use this type of constant to calculate all the parameters in the basis of this constant.


For example now in the basis of "k" we can write the bullet has a wavelength of approximately picometer.