Answer:
λ = 1.61x10⁻¹⁰ m
Explanation:
Broglie's wave equation is the following:
[tex] \lambda = \frac{h}{mv} [/tex]
Where:
λ: is the wavelength of the electron
h: is the Plank constant = 6.62x10⁻³⁴ J.s
m: is the mass of the electron = 9.11x10⁻³¹ kg
v: is the speed of the electron = 4.5x10⁶ m/s
The wavelength of the electron is:
[tex] \lambda = \frac{h}{mv} = \frac{6.62 \cdot 10^{-34 J.s}}{9.11 \cdot 10^{-31} kg*4.5 \cdot 10^{6} m/s} = 1.61 \cdot 10^{-10} m [/tex]
Therefore, the wavelength of an electron moving at 4.5 x 10⁶ m/s is 1.61x10⁻¹⁰ m.
I hope it helps you!
The wavelength of the electron is equal to 1.61 Angstrom
According to de Broglie's Wave Equation ,
[tex]\rm \lambda = \dfrac{h}{mv}[/tex]
where:
h=6.62607004×10⁻³⁴J⋅s is Planck's constant.
m = ⁻³¹kg is the rest mass of an electron.
v is the velocity it moves, in m/s
It is given that v = 4.5 x 10⁶ m/s
[tex]\rm \lambda = \dfrac{6.62607004×10^{-34}}{9.10938356\times10^{-31}\times 4.5\times10^{6}}}[/tex]
[tex]\rm \lambda = 1.61 \times 10^{-10} \\[/tex] m = 1.61 A°
Therefore the wavelength of the electron is equal to 1.61 Angstrom.
To know more about De Broglie's equation
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