Problem 173
Question
Calculate the wavelength (in monometer) associated with a proton moving at \(1.0 \times 10^{3} \mathrm{~ms}^{-1}\) (Mass of proton \(=1.67 \times 10^{-27} \mathrm{~kg}\) and \(\left.\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\right): \quad[\mathbf]\) (a) \(0.40 \mathrm{~nm}\) (b) \(2.5 \mathrm{~nm}\) (c) \(14.0 \mathrm{~nm}\) (d) \(0.032 \mathrm{~nm}\)
Step-by-Step Solution
Verified Answer
The wavelength is approximately 0.40 nm, corresponding to option (a).
1Step 1: Identify the formula required
To find the wavelength of a particle, we use the de Broglie wavelength formula: \( \lambda = \frac{h}{mv} \), where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.
2Step 2: Substitute the given values
Using the formula for de Broglie wavelength, substitute the given values: \( h = 6.63 \times 10^{-34} \, \text{Js} \), \( m = 1.67 \times 10^{-27} \, \text{kg} \), and \( v = 1.0 \times 10^{3} \, \text{ms}^{-1} \). The equation becomes: \[ \lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.0 \times 10^{3}} \]
3Step 3: Perform the calculation
Calculate the value of the denominator: \( m \times v = 1.67 \times 10^{-27} \times 1.0 \times 10^{3} = 1.67 \times 10^{-24} \)Now divide Planck's constant by this result to get the wavelength:\[ \lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-24}} = 3.97 \times 10^{-10} \, \text{m} \]
4Step 4: Convert units to nanometers
Since the problem asks for the wavelength in nanometers, convert from meters to nanometers (1 nanometer = \( 10^{-9} \) meters):\[ 3.97 \times 10^{-10} \, \text{m} = 0.397 \, \text{nm} \]
5Step 5: Choose the correct answer
Compare the calculated wavelength, \(0.397 \, \text{nm}\), to the options provided:(a) \(0.40 \, \text{nm}\)(b) \(2.5 \, \text{nm}\)(c) \(14.0 \, \text{nm}\)(d) \(0.032 \, \text{nm}\)The closest answer is (a) \(0.40 \, \text{nm}\).
Key Concepts
Planck's ConstantProton MassVelocity of a Particle
Planck's Constant
Planck's constant, often symbolized by \( h \), is a fundamental quantity in quantum physics. It plays a crucial role in the theories concerning the behavior of matter at microscopic scales. Planck's constant has the value \( 6.63 \times 10^{-34} \mathrm{~Js} \) and is a cornerstone in the equation for the de Broglie wavelength:
- It relates the energy of a photon to the frequency of its electromagnetic wave using the formula \( E = h u \), where \( u \) is the frequency.
- In the context of wave-particle duality, it allows one to calculate the wavelength of a particle, linking wave characteristics with particle mechanics.
Proton Mass
The proton mass is a critical component in calculating the de Broglie wavelength of particles like protons and electrons. It's approximately \( 1.67 \times 10^{-27} \mathrm{~kg} \). Understanding proton mass helps in:
- Determining how massive particles, compared to lighter ones like electrons, have significantly shorter wavelengths at the same velocity, thus experiencing reduced wave-like behaviors.
- Providing necessary context for how the mass impacts wavelength, given that a larger mass requires less wavelength for a constant velocity, as shown in the equation \( \lambda = \frac{h}{mv} \).
Velocity of a Particle
The velocity of a particle is a fundamental parameter in determining its de Broglie wavelength. Defined in meters per second \( \mathrm{m/s} \), velocity is a measure of how fast and in which direction a particle is moving. It greatly influences the computation of wavelength because:
- The velocity of the particle, alongside its mass, directly affects the wavelength. With higher velocities, particles like protons have shorter wavelengths.
- This relationship underscores the concept of wave-particle duality, where faster-moving particles exhibit less noticeable wave characteristics.
- In our problem, a velocity of \( 1.0 \times 10^{3} \mathrm{~ms}^{-1} \) was provided to calculate the proton's wavelength accurately using de Broglie's formulation \( \lambda = \frac{h}{mv} \).
Other exercises in this chapter
Problem 171
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