What is the kinetic energy of an electron that has a wavelength of 0.21 nm?

Steps to Solve

1. Identify the de Broglie wavelength formula
   The de Broglie wavelength $\lambda$ of a particle is given by the formula:
   $$ \lambda = \frac{h}{p} $$
   where $h$ is Planck's constant ($6.626 \times 10^{-34} , \text{Js}$) and $p$ is the momentum of the particle.

2. Express momentum in terms of kinetic energy
   The momentum $p$ can be related to kinetic energy ($KE$) by the formula:
   $$ KE = \frac{p^2}{2m} $$
   where $m$ is the mass of the particle. We can solve for momentum:
   $$ p = \sqrt{2m \cdot KE} $$

3. Combine the formulas
   Substituting the expression of $p$ from the kinetic energy formula into the de Broglie wavelength formula:
   $$ \lambda = \frac{h}{\sqrt{2m \cdot KE}} $$
   Now we can rearrange this to solve for kinetic energy:
   $$ KE = \frac{h^2}{2m \lambda^2} $$

4. Plug in known values
   Use the mass of an electron ($m = 9.11 \times 10^{-31} , \text{kg}$) and the given wavelength $\lambda$ (make sure to convert to meters if necessary) to calculate the kinetic energy.

5. Calculate kinetic energy
   Finally, substitute the values into the equation:
   $$ KE = \frac{(6.626 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times \lambda^2} $$