Question

A nitrogen molecule of mass \(m=4.648 \cdot 10^{-26} \mathrm{~kg}\) has a speed of \(300.0 \mathrm{~m} / \mathrm{s}\) a) Determine its de Broglie wavelength. b) How far apart are the double slits if a beam of nitrogen molecules creates an interference pattern with fringes \(0.30 \mathrm{~cm}\) apart on a screen \(70.0 \mathrm{~cm}\) in front of the slits?

Short answer

Answer: a) The de Broglie wavelength of the nitrogen molecule is \(4.75 \times 10^{-11} m\). b) The distance between the double slits is \(7.75 \times 10^{-8} m\) or \(77.5 nm\).

Step-by-step solution

Calculate the momentum of the nitrogen molecule

We are given that the mass of the nitrogen molecule is \(m = 4.648 \times 10^{-26} kg\) and its speed is \(v = 300.0 m/s\). To find its momentum, we can use the formula: \(p = m \cdot v\) So, \(p = 4.648 \times 10^{-26} kg \cdot 300.0 m/s = 1.3944 \times 10^{-23} kg \cdot m/s\). Step 2: Determine the de Broglie wavelength of the nitrogen molecule

Determine the de Broglie wavelength

Now that we have the momentum, we can find the de Broglie wavelength using the formula: λ = h / p where h is Planck's constant, \(6.626 \times 10^{-34} Js\). So, \(\lambda = \frac{6.626 \times 10^{-34} Js}{1.3944 \times 10^{-23} kg \cdot m/s} = 4.75 \times 10^{-11} m\) a) The de Broglie wavelength of the nitrogen molecule is \(4.75 \times 10^{-11} m\). Step 3: Calculate the angular separation of the fringes

Calculate the angular separation of the fringes

The fringes are given to be \(0.30 cm\) apart on the screen \(70 cm\) in front of the slits. To find the angular separation θ, we can use the small-angle approximation formula: \(\tan \theta \approx \theta = \frac{y}{L}\) where y is the fringe separation (0.30 cm) and L is the distance to the screen (70 cm). Converting the separation to meters, we get: \(\theta = \frac{0.003}{0.7} = 0.004286\) Step 4: Determine the distance between the double slits

Determine the distance between the double slits

Now, we can find the distance between the double slits using the formula: \(d = \frac{\lambda \cdot L}{n \cdot \theta}\) Since we have a first-order fringe (n=1), the formula becomes: \(d = \frac{4.75 \times 10^{-11} m \cdot 0.7 m}{0.004286} = 7.75 \times 10^{-8} m\) b) The distance between the double slits is \(7.75 \times 10^{-8} m\) or \(77.5 nm\).