How to calculate wavelength of de broglie that electron has been accelerated from rest through a potential difference of 1keV
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when the electron has been accelerated through a potential difference of 1KV the kinetic energy it develops is- 1000V X e=1.602 X 10 -19+3 =1.602 X 10 -16 J.
YOU MAY NOW USE THE RELATION p 2/2m=K.E
i.e
p 2=2 X 9.1 X 10 -31 X1.602 X 10 -16 =29.15 X 10 -47
p=17.07 X 10 -24 kg m s -1
lamda=h/p=6.626 X 10 -34 /17.07 X 10 -24 =0.388 X 10 -10 m=3.8 X 10 -11 m
ALTER:
YOU MAY ALSO USE THE RELATION :
LAMBDA=1.226x10 -9 /(V) 1/2 =1.226 X 10 -9 /(1000) 1/2 =3.8X10 -11 m
when the electron has been accelerated through a potential difference of 1KV the kinetic energy it develops is- 1000V X e=1.602 X 10 -19+3 =1.602 X 10 -16 J.
YOU MAY NOW USE THE RELATION p 2/2m=K.E
i.e
p 2=2 X 9.1 X 10 -31 X1.602 X 10 -16 =29.15 X 10 -47
p=17.07 X 10 -24 kg m s -1
lamda=h/p=6.626 X 10 -34 /17.07 X 10 -24 =0.388 X 10 -10 m=3.8 X 10 -11 m
ALTER:
YOU MAY ALSO USE THE RELATION :
LAMBDA=1.226x10 -9 /(V) 1/2 =1.226 X 10 -9 /(1000) 1/2 =3.8X10 -11 m
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