Answer :
Answer:
The electron has a wavelength of 1448 nanometers.
Explanation:
The wavelength of the electron can be determined by means of the de Broglie wavelength.
[tex]\lambda = \frac{h}{p}[/tex] (1)
Where h is the Planck's constant and p is the momentum
[tex]\lambda = \frac{h}{mv}[/tex] (2)
The velocity can be found by means of the kinetic energy equation.
[tex]KE = \frac{1}{2}mv^{2}[/tex]
[tex]2KE = mv^{2}[/tex]
[tex]v^{2} = \frac{2KE}{m}[/tex]
[tex]v = \sqrt{\frac{2KE}{m}}[/tex] (3)
Therefore, equation 3 can be replaced in equation 2
[tex]\lambda = \frac{h}{\sqrt{2mKE}}[/tex] (4)
Where m is the mass of the electron and KE its kinetic energy.
[tex]\lambda = \frac{6.624x10^{-34} J.s}{\sqrt{2(9.1x10^{-31}Kg)(1.15x10^{-19} J)}}[/tex]
But [tex]1J = Kg.m^{2}/s^{2}[/tex]
[tex]\lambda = \frac{6.624x10^{-34} Kg.m^{2}/s^{2}.s}{\sqrt{2.093x10^{-49}Kg^{2}.m^{2}/s^{2}}}[/tex]
[tex]\lambda = \frac{6.624x10^{-34} Kg.m^{2}/s^{2}.s}{4.574x10^{-25}Kg.m/s}[/tex]
[tex]\lambda = 1.448x10^{-9}m[/tex]
Finally, the wavelength can be expressed in terms of nanometers:
[tex]\lambda = 1.448x10^{-9}m .\frac{1x10^{9}nm}{1m}[/tex] ⇒ [tex]1448nm[/tex]
Hence, the electron has a wavelength of 1448 nanometers.
The wavelength of an electron that has a kinetic energy of 1.15 × 10^{-19} J is 1.45 nm.
From de-Broglie equation, the wavelength of an electron is given by:
[tex]\lambda=\frac{h}{\sqrt{2m(K.E)} }[/tex]
Where λ is the wavelength, h is the Planck constant = 6.626 * 10⁻34 J.s, m is the mass of electron = 9.1 * 10⁻³¹ kg, K.E = kinetic energy = 1.15 × 10⁻¹⁹ J.
Hence, substituting:
[tex]\lambda=\frac{h}{\sqrt{2m(K.E)} }\\\\\\\lambda=\frac{6.626*10^{-34}}{\sqrt{2*9.1*10^{-31}*1.15*10^{-19}} } \\\\\\\lambda=1.45\ nm[/tex]
The wavelength of an electron that has a kinetic energy of 1.15 × 10^{-19} J is 1.45 nm.
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