Question

Be sure to answer all parts. A baseball pitcher's fastballs have been clocked at about 97 mph (1 mile = 1609 m). (a) Calculate the wavelength of a 0.148−kg baseball (in nm) at this speed. × 10 nm (Enter your answer in scientific notation) (b) What is the wavelength of a hydrogen atom at the same speed? nm

Answer 1

Answer:

a) The wavelength of the baseball is $1.033\times 10^{-25} nm$.

b) 9.131 nm is the wavelength of a hydrogen atom at the 43.35 m/s.

Explanation:

Velocity of the baseball = v = 97 mile/hour

1 mile = 1609 meter

1 hour = 3600 seconds

$v =\frac{97\times 1609 m}{3600 s}=43.35 m/s$

Mass of baseball = m = 0.148 kg

Wavelength of the baseball: $\lambda$

$\lambda =\frac{h}{mv}$     De Broglie wavelength

h =Planck's constant

$=\frac{6.626\times 10^{-34} Js}{0.148 kg\times 43.35 m/s}$

$\lambda =1.033\times 10^{-34} m$

$1 m=10^9 nm$

$\lambda = 1.033\times 10^{-25} nm$

The wavelength of the baseball is $1.033\times 10^{-25} nm$.

b)

Mass of the hydrogen atom = $m=1.674\times 10^{-27} kg$

Velocity of hydrogen atom = u = 43.35 m/s

$\lambda =\frac{h}{mv}$    De Broglie wavelength

$=\frac{6.626\times 10^{-34} Js}{1.674\times 10^{-27} kg\times 43.35 m/s}$

$\lambda =9.131\times 10^{-9} m$

$1 m=10^9 nm$

$\lambda =9.131 nm$

9.131 nm is the wavelength of a hydrogen atom at the 43.35 m/s.