1. Calculate the de Broglie wavelength in picometers (1 picometer =10^−12 meters) of an electron that has kinetic energy of 10 keV. The mass of an electron is 9.11 x 10^−31 kg
SOMEONE POSTED:
KE=10⁴•1.6•10⁻¹⁹ =1.6•10 ⁻¹⁵J
KE=mv²/2 =m²v²/2m =p²/2m =>
p=sqrt(2m•KE)
λ=h/p =h/sqrt((2m•KE)=
=6.63•10⁻³⁴/sqrt(2•9.1•10 ⁻³¹•1.6•10 ⁻¹⁵)=
=1.22•10⁻¹¹ m=0.122 pm
Still wrong answer...anyone have another answer?
My mistake in the last line
=1.22•10⁻¹¹ m=12.2 pm
To calculate the de Broglie wavelength of an electron, you'll need to use the equation λ = h / p, where λ is the wavelength, h is Planck's constant (6.63 x 10^-34 J·s), and p is the momentum of the electron. The formula for momentum is given by p = √(2m·KE), where m is the mass of the electron (9.11 x 10^-31 kg) and KE is the kinetic energy.
Let's go through the calculations step by step:
1. Convert the kinetic energy from kiloelectron volts (keV) to joules (J). Remember that 1 electron volt (eV) is equal to 1.6 x 10^-19 J. So, 10 keV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J.
2. Use the formula p = √(2m·KE) to calculate the momentum:
p = √(2 x 9.11 x 10^-31 kg x 1.6 x 10^-18 J)
p ≈ √(2 x 1.4576 x 10^-48 kg·J)
p ≈ √2.9152 x 10^-48 kg·J
p ≈ 1.71 x 10^-24 kg·m/s
3. Now, you can calculate the de Broglie wavelength λ = h / p:
λ = 6.63 x 10^-34 J·s / (1.71 x 10^-24 kg·m/s)
λ ≈ 3.87 x 10^-10 m
4. Finally, convert the wavelength to picometers (pm):
1 m = 10^12 pm, so λ ≈ 3.87 x 10^-10 m x 10^12 pm/m
λ ≈ 3.87 x 10^2 pm
λ ≈ 387 pm
Therefore, the de Broglie wavelength of the electron with a kinetic energy of 10 keV is approximately 387 picometers (pm).