Question #46f0c

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Question #46f0c

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Answer 1 · 2016-05-22T16:51:36

$4.69 \times 10^{14}$ $4.69times10^14$ MeV and $λ = 2.24 \times 10^{- 34} m$ $lamda=2.24times10^-34m$

Explanation:

a quick search in the internet suggests that Federer's average first serve at Wimbledon was about 115 mph in 2016 which is about 51 m/s.

The mass of a tennis ball is about 58g (0.058kg).

So the kinetic energy would be $K E = \frac{1}{2} m v^{2} = \frac{1}{2} \cdot 0.058 \cdot 51^{2} = 75 J$ $KE=1/2mv^2=1/2*0.058*51^2=75J$

An electron volt = $1.6 \times 10^{- 19} J$ $1.6 times 10^-19 J$ and a MeV = $1.6 \times 10^{- 13} J$ $1.6times 10^-13J$

So in MeV, the energy would be $\frac{75}{1.6 \times 10^{- 13}} = 4.69 \times 10^{14} M e V$ $75/(1.6times10^-13)=4.69times10^14MeV$

The de Boglie wavelength, $λ$ $lamda$ is given by $λ = \frac{h}{m v}$ $lamda=h/(mv)$
So $λ = \frac{6.63 \times 10^{- 34}}{0.058 \cdot 51} = 2.24 \times 10^{- 34} m$ $lamda=(6.63times 10^-34)/(0.058*51)=2.24times10^-34m$

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Question #46f0c

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