The scattering of helium atoms by a solid can be used to gain information about the lattice of atoms, provided the wavelength of the helium atoms is comparable to the dimensions of the lattice. What is the wavelength, in picometers, of a helium-4 atom with a kinetic energy of 0.10 eV (Mass of ⁴He = 4.0026 amu; 1 amu = 1.67 * 10^-24 g; 1 eV = 1.602 * 10^-19 J; h = 6.626 * 10^-34 J*s) A) 55 pm B) 45 pm C) 35 pm D) 25 pm E) None of the above

Question

Quizplus · Accepted Answer

The de Broglie wavelength of a particle with mass m and velocity v is given by:

λ = h / mv

Since the helium atom has a kinetic energy of 0.10 eV, we can find its velocity using:

KE = (1/2)mv^2

0.10 eV = (1/2)(4.0026 amu)(v^2)

v = 5290 m/s

Now we can calculate the de Broglie wavelength using:

λ = h / mv

λ = (6.626 x 10^-34 J·s) / (4.0026 x 1.67 x 10^-27 kg) / (5290 m/s)

λ = 45 pm

Therefore, the answer is B.

The Scattering of Helium Atoms by a Solid Can Be