The de Broglie wavelength (λ) of a particle is related to its momentum (p) by the equation:
λ = h / p
Where λ is the de Broglie wavelength, h is the Planck's constant (approximately 6.626 x 10^(-34) J·s), and p is the momentum of the particle.
To calculate the kinetic energy (KE) of an electron, we need to find its momentum first. The momentum of an electron is given by:
p = mv
Where m is the mass of the electron (approximately 9.10938356 x 10^(-31) kg) and v is the velocity of the electron.
Since the electron is not accelerated and we are given its de Broglie wavelength, we can use the equation:
λ = h / mv
Rearranging the equation, we can solve for the velocity (v):
v = h / (mλ)
Now we have the velocity of the electron, and we can calculate its kinetic energy using the formula:
KE = (1/2)mv^2
Substituting the value of v we obtained earlier, we can calculate the kinetic energy. Let's calculate it:
λ = 200 pm = 200 x 10^(-12) m m = 9.10938356 x 10^(-31) kg h = 6.626 x 10^(-34) J·s
First, calculate the velocity:
v = h / (mλ) v = (6.626 x 10^(-34) J·s) / ((9.10938356 x 10^(-31) kg)(200 x 10^(-12) m)) v ≈ 3.319 x 10^6 m/s
Next, calculate the kinetic energy:
KE = (1/2)mv^2 KE = (1/2)(9.10938356 x 10^(-31) kg)(3.319 x 10^6 m/s)^2 KE ≈ 4.498 x 10^(-18) J
Therefore, the kinetic energy of an electron with a de Broglie wavelength of 200 pm is approximately 4.498 x 10^(-18) J.