Alpha particles (m = 6.6 * 10-27 kg,q = +2e)are accelerated through a potential difference of 11,100V,what is their de Broglie wavelength? A)1.9 * 10^-49 m B)1.5 * 10^-15 m C)9.7 * 10^-14 m D)4.5 * 10^-12 m E)9.0 * 10^-12 m

Question

Quizplus · Accepted Answer

The de Broglie wavelength of a particle is given by the equation λ = h/p, where h is Planck's constant and p is the momentum of the particle. 
The momentum of an alpha particle can be calculated as follows: 
p = √(2mK), where m is the mass of the alpha particle, K is the kinetic energy gained by the particle due to the potential difference, and the factor of 2 comes from the fact that the alpha particle is doubly charged (q = +2e). 
Substituting the given values, we get: 
K = qV = (2)(1.6 × 10^-19 C)(11,100 V) = 3.52 × 10^-15 J 
p = √(2mK) = √(2)(6.6 × 10^-27 kg)(3.52 × 10^-15 J) = 2.82 × 10^-21 kg m/s 
Finally, we can calculate the de Broglie wavelength: 
λ = h/p = (6.63 × 10^-34 J s)/(2.82 × 10^-21 kg m/s) = 9.7 × 10^-14 m = 9.7 × 10^-10 nm = 9.7 Å 
Thus, the best choice is C.

Alpha Particles (M = 6