Respuesta :
Question in incomplete, complete question is:
Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of [tex]4.71\times 10^{-15}J[/tex] . What is the de Broglie wavelength of this electron (Ek = ½mv²)?
Answer:
[tex] 6.762\times 10^{-12} m[/tex] is the de Broglie wavelength of this electron.
Explanation:
To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:
[tex]\lambda=\frac{h}{\sqrt{2mE_k}}[/tex]
where,
= De-Broglie's wavelength = ?
h = Planck's constant = [tex]6.624\times 10^{-34}Js[/tex]
m = mass of beta particle = [tex] 9.1094\times 10^{-31} kg[/tex]
[tex]E_k[/tex] = kinetic energy of the particle = [tex]4.71\times 10^{-15}J[/tex]
Putting values in above equation, we get:
[tex]\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}[/tex]
[tex]\lambda = 6.762\times 10^{-12} m[/tex]
[tex] 6.762\times 10^{-12} m[/tex] is the de Broglie wavelength of this electron.
