Question

An alpha particle (mass = 6.6 x 10²⁴ g) emitted by radium travels at 3.4 x 10⁷ ± 0.1 x 10⁷ mi/h.

Answer 1

Answer:

For a: The De-Broglie wavelength of the given particle is $2\times 10^{-10}m$

For b: The uncertainty in the position is $3.56\times 10^{-14}m$

Explanation:

• For a:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{mv}$

where,

$\lambda$ = De-Broglie's wavelength = ?

h = Planck's constant =

m = mass of the alpha particle =

v = velocity of alpha particle = $3.4\times 10^7mi/h=1.52\times 10^7m/s$  (Conversion factors used:  1 mile = 1609.34 m  &  1 hr = 3600 s)

Putting values in above equation, we get:

$\lambda=\frac{6.6\times 10^{-34}Js}{6.6\times 10^{-27}kg\times 1.52\times 10^7m/s}\\\\\lambda=6.58\times 10^{-15}m$

Hence, the De-Broglie wavelength of the given particle is $2\times 10^{-10}m$

• For b:

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p\geq \frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = ?

$\Delta p$ = uncertainty in momentum  = $m\Delta v$

m = mass of the alpha particle = $6.6\times 10^{-27}kg$

$\Delta v$ = uncertainty in speed = $0.1\times 10^7mi/hr=4.47\times 10^5m/s$

h = Planck's constant = $6.6\times 10^{-34}kgm^2/s$

Putting values in above equation, we get:

$\Delta x=\frac{6.6\times 10^{-34}kgm^2/s}{2\times 3.14 \times 6.6\times 10^{-27}\times kg 4.47\times 10^{5}m/s}\\\\\Delta x=3.56\times 10^{-14}m$

Hence, the uncertainty in the position is $3.56\times 10^{-14}m$