Question

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 4.71 × 10−15 J. What is the de Broglie wavelength of this electron (Ek = mu2/2)?

Answer 1

Answer : The de-Broglie wavelength of this electron, $0.101\AA$

Explanation :

The formula used for kinetic energy is,

$K.E=\frac{1}{2}mv^2$        ..........(1)

According to de-Broglie, the expression for wavelength is,

$\lambda=\frac{h}{mv}$

or,

$v=\frac{h}{m\lambda}$      ...........(2)

Now put the equation (2) in equation (1), we get:

$\lambda=\frac{h}{\sqrt{2\times m\times K.E}}$  ...........(3)

where,

$\lambda$ = wavelength = ?

h = Planck's constant = $6.626\times 10^{-34}Js$

m = mass of electron = $9.11\times 10^{-31}Kg$

K.E = kinetic energy = $4.71\times 10^{-15}J$

Now put all the given values in the above formula (3), we get:

$\lambda=\frac{6.626\times 10^{-34}Js}{\sqrt{2\times 9.11\times 10^{-31}Kg\times 4.71\times 10^{-15}J}}$

$\lambda=1.0115\times 10^{-11}m=0.101\AA$

conversion used : $(1\AA=10^{-10}m)$

Therefore, the de-Broglie wavelength of this electron, $0.101\AA$