Question

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, the ball will break through the strings and keep going. The racket is a potential-energy barrier whose height is the energy of the slowest string-breaking ball. Suppose that a 100 g tennis ball traveling at 220mph is just sufficient to break the 2.0-mm-thick strings. Estimate the probability that a 150mph ball will tunnel through the racket without breaking the strings.

Give your answer as a power of 10 rather than a power of e.

Answer 1

Basically the ball is described by a wave function psi. Since it needs to go through a barrier of potential the wave function will decrease exponentially in the barrier with the barrier thickness x.

Psi(x) = Psi(0)*exp(-kx)

k is the wave number so that (kx) is adimensional.
k =1/lambda = P/h_bar= sqrt(2*m*(U-E))/ h_bar
U is the height of the barrier, E is the kinetic energy of the incoming ball, m is the ball mass

(it is eactly like when an electron of energy E tunnel through a potential barrier U)

V1 =150 mph =67.0 m/s

V2= 220 mph =98.35 m/s

(conversions done by google)

The height of the potential barrier is equal to the maximum energy of the incoming ball that breaks the strings

U = mv2^2/2 = 0.1*98.35^2 /2 =483.6 J

Incoming ball has a kinetic energy

Ek =mv1^2/2 = 0.1*67^2/2 =224.45 J

U - E =259.15 J

The wavenumber is

k =sqrt(2*0.1*259.15)/10^-34 =7.2*10^34

The probability of tunneling is just (d=2 mm is the thickness of the barrier)
P = |psi(d)/psi(0) |^2 = exp(-2*k*d) =exp(-2*7.2*10^34*2*10^-3) =exp(-28.8*10^31)

Now we know that exp(2.303) =10 so that

P = e^(2.303)^(-12.5*10^31) =10^(-12.5*10^31)

Check link http://en.wikipedia.org/wiki/Scanning_tunneling_microscope#Principle_of_operation

(to verify the probability of tunneling)