Question

You decide to roll a 0.15-kg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelength. A.If you roll it at 1.2×10−3 m/s , what is its wavelength? B.How does this compare with the de Broglie wavelength of the high-speed electron that strikes the back face of one of the early models of a TV screen at 1/10 the speed of light (2.4×10−11m)?

Answer 1

A)

m = mass of the ball = 0.15 kg

v = speed of the ball = 1.2 x 10^-3 m/s

wavelength is given using de Broglie's principle as

= h/(mv)

= (6.63 x 10^-34)/((0.15) (1.2 x 10^-3))

= 3.7 x 10^-30 m

b)

m_e = mass of the electron = 9.1 x 10^-31 kg

v_e = speed of the electron = (3 x 10⁸)/10 = 3 x 10⁷ m/s

wavelength is given using de Broglie's principle as

_e = h/(m_ev_e)

_e  = (6.63 x 10^-34)/((9.1 x 10^-31) (3 x 10⁷))

_e  = 2.4 x 10^-11 m

Ratio is given as

/_e  = 3.7 x 10^-30 /( 2.4 x 10^-11)

/_e  = 1.5 x 10^-19