Question

A baseball pitcher's fastballs have been clocked at about 91 mph (1 mile = 1609 m). (a) Calculate the wavelength of a 0.143-kg baseball (in nm) at this speed. {Enteryour answer in scientific notation) (b) What is the wavelength of a hydrogen atom at the same speed? nm

Answer 1

Here we know that De Broglie equation that

$\\\lambda =\frac{h}{mv}$

M = mass

V = velocity

Here we have

M = 0.143 kg

$h=plank's\quad constant=6.626\times 10^{-34}JS$

V = 91 mph = 40.68 m/s (1 mile = 1069 m)

Now apply

$\\\lambda =\frac{h}{mv}$

$\\\lambda =\frac{6.626\times 10^{-34}Js}{0.143Kg\times 40.68m/s}\\ \lambda =\frac{6.626\times 10^{-34}Js}{5.81Kg\times m/s}\\ \lambda =1.14\times 10^{-34} m\\ \lambda =1.14\times 10^{-34} \times 10^9 nm\\ \lambda =1.14\times 10^{-25}nm\\$

b) Mass of hydrogen $1.67\times 10^{-27}Kg$

Velocity is 40.68 m/s

Now apply the de Broglie's equation

$\\\lambda =\frac{h}{mv}$

$\\\lambda =\frac{6.626\times 10^{-34}Js}{1.67\times 10^{-27}Kg\times 40.68m/s}\\ \lambda =\frac{6.626\times 10^{-34}Js}{67.93\times 10^{-27}Kg\times m/s}\\ \lambda =0.09\times 10^{-7} m\\ \lambda =9\times 10^{-9} \times 10^9 nm\\ \lambda =9nm\\$

Hence the wavelength of Hydrogen is 9 nm.

Answer.