Question

Assuming that the smallest measurable wavelength in an experiment is 0.350 fm (femtometers), what is the maximum mass of an object traveling at 465 m·s–1 for which the de Broglie wavelength is observable?

Answer 1

Answer 2

Let's break down this problem step-by-step:

1. Understand the de Broglie Wavelength

The de Broglie wavelength (λ) of a particle is given by the formula:

λ = h / (mv)

where:

• λ is the de Broglie wavelength

• h is Planck's constant (approximately 6.626 x 10⁻³⁴ J·s)

• m is the mass of the particle

• v is the velocity of the particle

2. Identify the Given Information

• Smallest measurable wavelength (λ) = 0.350 fm = 0.350 x 10⁻¹⁵ m

• Velocity (v) = 465 m/s

3. Determine the Maximum Mass

We want to find the maximum mass (m) for which the de Broglie wavelength is observable.  Since we are given the smallest measurable wavelength, we need to find the mass that corresponds to this wavelength.

Rearrange the de Broglie wavelength formula to solve for mass (m):

m = h / (λv)

4. Plug in the Values and Calculate

m = (6.626 x 10⁻³⁴ J·s) / (0.350 x 10⁻¹⁵ m * 465 m/s)

m ≈ 4.06 x 10⁻²² kg

Answer:

The maximum mass of the object for which the de Broglie wavelength is observable is approximately 4.06 x 10⁻²² kg.

Answer 3

we'll use the de Broglie wavelength formula:

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

Where:

• $\lambda$ is the wavelength (0.350 fm = $0.350\times 10^{-15}$ m).

• $h$ is Planck's constant ($6.626\times 10^{-34}$ J·s).

• $m$ is the mass of the object (in kg).

• $v$ is the velocity (465 m/s).

 

1. Rearrange the formula to solve for mass ($m$):

   $$m=\frac{h}{\lambda v}$$

2. Plug in the given values:

   $$m=\frac{6.626\times 10^{-34}}{(0.350\times 10^{-15})\times 465}$$

3. Simplify the denominator:

   $$\lambda v=0.350\times 10^{-15}\times 465=162.75\times 10^{-15}=1.6275\times 10^{-13}{\text{ m}}^2\mathrm{/}\text{s}$$

4. $$=162.75\times 10^{-15}$$

5. $$=1.6275\times 10^{-13}{\text{ m}}^2/\text{s}$$

6. Calculate the mass:

   $$m=\frac{6.626\times 10^{-34}}{1.6275\times 10^{-13}}\approx 4.07\times 10^{-21}\text{ kg}$$

7. 
   

8. $$\approx 4.07\times 10^{-21}\text{ kg}$$

9. Convert to more familiar units (optional):

   $$4.07\times 10^{-21}\text{ kg}\approx 4.07\times 10^{-18}\text{ g}$$

 Answer is :

The maximum mass of an object traveling at 465 m/s for which the de Broglie wavelength is observable (with a minimum measurable wavelength of 0.350 fm) is approximately 4.07 × 10⁻²¹ kg .