Everything Science Grade 12

12.2 CHAPTER 12. WAVE NATURE OF MATTER

λ =

h

mv

=

6 , 63 × 10 −^34 J· s

(0,150 kg)(40 m· s−^1 )

= 1, 11 × 10 −^34 m

This wavelength is considerably smaller thanthe diameter of a

proton which is approximately 10 −^15 m. Hence the wave-like

properties of this cricketball are too small to beobserved.

Example 2: The de Broglie wavelength of an electron

QUESTION

Calculate the de Brogliewavelength of an electron moving at 40 m·s−^1.

SOLUTION

Step 1 : Determine what is required and how to approach the problem

We are required to calculate the de Broglie wavelength of an

electron given its speed. We can do this by using:

λ =

h

mv

Step 2 : Determine what is given

We are given:

• The velocity of the electron v = 40 m· s−^1
  and we know:


• The mass of the electron me= 9, 11 × 10 −^31 kg


• Planck’s constant h = 6, 63 × 10 −^34 J· s

Step 3 : Calculate the de Broglie wavelength

λ =

h

mv

=

6 , 63 × 10 −^34 J· s

(9, 11 × 10 −^31 kg)(40 m· s−^1 )

= 1, 82 × 10 −^5 m

= 0,0182 mm