1.

An electron, a proton and an alpha particle have K.E. of 16E, 16E,4E respectively?

Answer»

deBroglie wavelength is λ = h/(mv)

Kinetic energy is K = ½mv² 

so v = √(2K/m)

λ = h/(m√(2K/m)) 

= h/(√(2Km²/m)) 

= h/(√(2Km) 

= (h/√2)(1/√(Km))

So λ is proportional to 1/√(Km)

For the electron m = 1/2000 amu approx and K =16E. 

1/√(Km) = 1/√(16E x (1/2000)) = 11.2/√E approx

For the proton m = 1 amu approx and K =16E. 

1/√(Km) = 1/√(16E x 1) = 0.25/√E

For the alpha particle m = 4 amu approx and K =4E 

1/√(Km) = 1/√(4E x 4) = 0.25/√E

So the order of their wavelengths (shortest to longest) is: 

proton and alpha particle equal, electron shortest.

If you inspect the expression 

1/√(16E x (1/2000)) 

1/√(16E x 1) 

1/√(4E x 4) y 

The quantitative ratio of their wavelengths is: λp:λα:λe = 0.25:0.25:11.2.


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