Asked by Kushin Dunos
1. Calculate the de Broglie wavelength in picometers (1 picometer =10^−12 meters) of an electron that has kinetic energy of 10 keV. The mass of an electron is 9.11 x 10^−31 kg
SOMEONE POSTED:
KE=10⁴•1.6•10⁻¹⁹ =1.6•10 ⁻¹⁵J
KE=mv²/2 =m²v²/2m =p²/2m =>
p=sqrt(2m•KE)
λ=h/p =h/sqrt((2m•KE)=
=6.63•10⁻³⁴/sqrt(2•9.1•10 ⁻³¹•1.6•10 ⁻¹⁵)=
=1.22•10⁻¹¹ m=0.122 pm
Still wrong answer...anyone have another answer?
SOMEONE POSTED:
KE=10⁴•1.6•10⁻¹⁹ =1.6•10 ⁻¹⁵J
KE=mv²/2 =m²v²/2m =p²/2m =>
p=sqrt(2m•KE)
λ=h/p =h/sqrt((2m•KE)=
=6.63•10⁻³⁴/sqrt(2•9.1•10 ⁻³¹•1.6•10 ⁻¹⁵)=
=1.22•10⁻¹¹ m=0.122 pm
Still wrong answer...anyone have another answer?
Answers
Answered by
Elena
My mistake in the last line
=1.22•10⁻¹¹ m=12.2 pm
=1.22•10⁻¹¹ m=12.2 pm
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