An electron has a de Broglie wavelength of 240nm, What is the speed of the electron?

Question

An electron has a de Broglie wavelength of 240nm, What is the speed of the electron? I dont get how to do this sompel please help.

Answer 1

wavelength = h/mv v= speed v= h/m*wavelength

v= 6.62*10^-34/9*10^-31*240*10^-9=3065m/s

Answer 2

λ = h/mv x √ ( 1-v²/c²)

where λ = debroglie wavelength = 240 nm =2400 x 10‾¹º m
h = plancks constant = 0.6626 x 10‾³³ J*s = 0.6626 x 10‾³³ kgm²/s
m = electron rest mass = 9.109 x 10‾³¹ kg
c = velocity of light = .03 x 10¹º m/s
c² = .0009 x 10²º m²/s²

so....

λ = h/m x √ [ ( 1-v²/c²)/v²]
λ * m/h = √ [ ( 1-v²/c²)/v²]
(λ * m/h )² = [ ( 1-v²/c²)/v²]
(λ * m/h )² = ( 1-v²/c²)/v²
v² = ( 1-v²/c²)/(λ * m/h )² = 1/(λ * m/h )² - (v²/c²)/(λ * m/h )²
v² + (v²/c²)/(λ * m/h )² = 1/(λ * m/h )²
v² * [ 1 + (1/c²)/(λ * m/h )²] = 1/(λ * m/h )²

v = √ {(1/(λ * m/h )²) / [ 1 + (1/c²)/(λ * m/h )²] }

v = √{(1/(2400 x 10‾¹º m * 9.109 x 10‾³¹ kg / (0.6626 x 10‾³³ kgm²/s) )²) / [ 1 + (1/(.0009 x 10²º m²/s²))/(2400 x 10‾¹º m * 9.109 x 10‾³¹ kg / (0.6626 x 10‾³³ kgm²/s) )²]

v = √{((1/(.330 x 10‾³ s/m )²) / [ 1 + (1/(.0009 x 10²º m²/s²) / (.330 x 10‾³ s/m )²] }

v = √{(9186 x 10³ m²/s²) / [ 1 + (1/(.0009 x 10²º m²/s²) / (1088 x 10‾¹º s²/m² )] }

v = √{(9186 x 10³ m²/s²) / [ 1 + (1.02 x 10‾¹º)] }

v = √{(9186 x 10³ m²/s²) } = 3030 m/s