Question

calculate the velocity and kinetic energy of an electron having wave length of 0.21 nm ​

Answer 1

Answer:

De-Broglie wavelength, λ=h/mv

λ

2

=

m

2

v

2

h

2

λ

2

=

2m(K.E.)

h

2

K.E.=

2mλ

2

h

2

=

2×9.1×10

−31

×(10

−9

)

2

(6.6×10

−34

)

2

=2.42×10

−19

J

=

1.6×10

−19

2.42×10

−19

eV

≈1.5 eV