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What is the de Broglie wavelength of a bullet of mass 0.075 kg traveling at 350 m/s? Planck's Constant is 6.63 × 10−34 J⋅s.
4.1 × 10−^35 m
2.5 × 10−^35 m
2.2 × 10−^34 m
1.7 × 10−^34 m

Respuesta :

Answer:

De Broglie wavelength of the bullet is given as

[tex]\lambda = 2.5 \times 10^{-35} m[/tex]

Explanation:

As per De Broglie hypothesis we know that the wavelength of De Broglie waves is the ratio of Plank's constant and momentum of the particle

here we know that the bullet mass is

m = 0.075 kg

speed of the bullet is given as

v = 350 m/s

Now we have

[tex]P = (0.075)(350)[/tex]

[tex]P = 26.25 kg m/s[/tex]

Now we know that

[tex]\lambda = \frac{h}{P}[/tex]

[tex]\lambda = \frac{6.63 \times 10^{-34}}{26.25}[/tex]

[tex]\lambda = 2.5 \times 10^{-35} m[/tex]

Cricetus

The de Broglie wavelength of a bullet will be "2.5 × 10⁻³⁵ m". To understand the calculation, check below.

Mass and Wavelength

According to the question,

Mass of bullet, m = 0.075 kg

Speed of bullet, v = 350 m/s

Planck's constant, h = 6.63 × 10⁻³⁴ J.s

We know the relation,

→ P = m × v

By substituting the values,

     = 0.075 × 350

     = 26.25 kg.m/s

hence,

The wavelength will be:

→ λ = [tex]\frac{h}{P}[/tex]

     = [tex]\frac{6.63\times 10^{-34}}{26.25}[/tex]

     = 2.5 × 10⁻³⁵ m

Thus the approach above is correct.

Find out more information about wavelength here:

https://brainly.com/question/5440536

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