Respuesta :
Answer: The de broglie wavelength is [tex]2.037 \times 10^{-34} m[/tex].
Explanation:
Calculate [tex]\lambda = \frac{h}{p}[/tex]as follows.
[tex]\lambda = \frac{h}{p}[/tex]
where,
h = plank's constant = [tex]6.6 \times 10^{-34} m^{2} kg/s[/tex]
p = momentum = [tex]mass \times velocity[/tex]
Putting the values in the formula as follows.
[tex]\lambda = \frac{h}{mass \times velocity}[/tex]
= [tex]\frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}[/tex]
= [tex]2.037 \times 10^{-34} m[/tex]
Thus, the de broglie wavelength is [tex]2.037 \times 10^{-34} m[/tex].
The value of de Broglie wavelength of golf ball is [tex]\boxed{2.045 \times {{10}^{ - 34}}{\text{ m}}}[/tex].
Further Explanation:
de Broglie wavelength:
The de Broglie equation is used to calculate the unknown value of wavelength from the known values of mass and velocity. It is specially applied to neutral atoms, elementary particles, and molecules. The de Broglie equation is as follows:
[tex]\lambda=\frac{h}{{mv}}[/tex] .......(1)
Here, m is the mass in kilogram, h is the Planck’s constant whose value is equal to [tex]6.626\times{10^{ - 34}}{\text{ J}}\cdot {\text{sec}}[/tex], [tex]\lambda[/tex] is the de Broglie wavelength in meters, and v is the velocity in meter per second.
The velocity of a golf ball is [tex]72{\text{ m/s}}[/tex].
The mass of a golf ball is [tex]45\,{\text{g}}[/tex].
Mass of golf ball in kilogram is calculated as follows:
[tex]\begin{aligned}{\text{Mass}}\left({{\text{kg}}}\right)&={\text{Mass}}\left( {\text{g}} \right)\left({\frac{{1{\text{ kg}}}}{{1000{\text{ g}}}}}\right)\\&=\left( {45\,{\text{g}}} \right)\left({\frac{{1{\text{ kg}}}}{{1000{\text{ g}}}}}\right)\\&=0.045\,{\text{kg}}\\\end{gathered}[/tex]
Substitute [tex]6.626\times {10^{ - 34}}{\text{ J}}\cdot{\text{sec}}[/tex] for h, [tex]0.045\;{\text{kg}}[/tex] for m, and [tex]72{\text{ m/s}}[/tex] for v in equation (1) to calculate the value of de Broglie wavelength [tex]\left(\lambda\right)[/tex].
[tex]\begin{aligned}\lambda&=\frac{h}{{mv}}\\&=\frac{{\left( {6.626 \times {{10}^{ - 34}}{\text{ J}}\cdot {\text{sec}}}\right)}}{{\left( {0.045\,{\text{kg}}}\right)\left({72{\text{ m/s}}}\right)}}\\&= 2.045 \times {10^{ - 34}}{\text{ m}}\\\end{gathered}[/tex]
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Atomic structure
Keywords: de Broglie equation, m, h, v, [tex]\lambda[/tex], wavelength, velocity, mass, plank’s constant, kilogram and golf ball.
