Question

7.42 What is the de Broglie wavelength (in nm) associ- iated with a 2.5-g Ping-Pong ball traveling 35 mph?

Answer 1

Given:

Mass of the Ping-Pong ball = 2.5 g

Speed of the Ping-Pong ball = 35 mph

STEP 1: Calculate the mass in SI units

m = 2.59

$m=2.5 \times 10^{-3} kg$   

( 19 = 10-3kg)

STEP 2: Calculate the speed in SI units

V = 35 X 0.447m/s

  $(\because 1mph = 0.447 m/s)$

STEP 3: Use the de- Broglie equation to calculate the wavelength of the ball

$\lambda =\frac{h}{mv}$   where h is the Planck's Constant

$\lambda =\frac{6.626\times 10^{-34}Js}{(2.5\times 10^{-3}kg)\times (15.645 m/s)}$

$\lambda =1.69 \times 10^{-32} m$

STEP 4: Convert meters into nanometers

$\because 1m=10^{9} nm$

$\therefore \lambda =1.69 \times 10^{-32} \times 10^{9} nm$

$\lambda =1.69 \times 10^{-23} nm$

Therefore, the wavelength associated with the Ping-Pong ball is $1.69 \times 10^{-23} nm$