a What is the de Broglie wavelength of a helium nucleus with
a. What is the de Broglie wavelength of a helium nucleus with an energy of 0.001 eV?
b. What is the de Broglie wavelength of a 45 g golf ball traveling at 10 mph?
Solution
a. De- broglie wavelenth,
E = h c / lambda
hc = 1240 eV - nm
0.001 eV = 1240 ev -nm / lambda
lambda = 1.24 x 10^6 nm = 1.24 x 10^-3 m Or 1.24 mm
(B) wavelength, lambda = h / p
p = m v
m = 45 x 10^-3 kg
v = 10 mph = 10 x 1609 m / 3600s = 4.47 m/s
= (6.626 x 10^-34) / (45x 10^-3) (4.47)
= 3.29 x 10^-33 m
