Problem 173 - A Complete Resource Book in Chemistry for JEE Main

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 \(\left.\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\right): \quad[\mathbf{2 0 0 9}]\) (a) \(0.40 \mathrm{~nm}\) (b) \(2.5 \mathrm{~nm}\) (c) \(14.0 \mathrm{~nm}\) (d) \(0.032 \mathrm{~nm}\)

Step-by-Step Solution

Answer

The wavelength of the proton is approximately 0.40 nm, corresponding to option (a).

Step 1: Identify Known Variables

We know the mass of the proton, \( m = 1.67 \times 10^{-27} \mathrm{~kg} \). The speed of the proton is \( v = 1.0 \times 10^{3} \mathrm{~ms}^{-1} \). Planck's constant is \( h = 6.63 \times 10^{-34} \mathrm{~Js} \).

Step 2: Understand De Broglie Wavelength Formula

The de Broglie wavelength formula is given by \( \lambda = \frac{h}{mv} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is the velocity of the particle.

Step 3: Substitute Values into the Formula

Substitute the given values into the de Broglie wavelength formula: \[ \lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.0 \times 10^{3}} \].

Step 4: Calculate the Wavelength

Calculate \( \lambda \): \[ \lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.0 \times 10^{3}} = 3.97 \times 10^{-10} \mathrm{~m} \].

Step 5: Convert Wavelength to Nanometers

To convert the wavelength from meters to nanometers (1 nm = \( 10^{-9} \) meters): \[ 3.97 \times 10^{-10} \mathrm{~m} = 0.397 \times 10^{-9} \mathrm{~m} = 0.397 \mathrm{~nm} \].

Step 6: Choose the Closest Answer Option

Find the closest answer option to the calculated wavelength, 0.397 nm. It is option (a) \(0.40 \mathrm{~nm}\).

Key Concepts

Proton

A proton is a fundamental particle that resides in the nucleus of an atom. Protons have a positive electric charge and play a critical role in defining the atomic number of an element. They have a relatively small mass, about 1.67 × 10^{-27} kg, which makes them incredibly lightweight compared to other particles like neutrons and especially massive objects like planets.
Understanding the properties of protons is crucial in physics, especially when discussing particles at an atomic level. Protons, along with neutrons, contribute to the mass of an atom, despite being much smaller than everyday objects.
In the realm of quantum mechanics, protons can also demonstrate wave-like behavior, which means they can be described by a wavelength. This is particularly interesting when examining their motion and interaction in particle physics, such as when you are calculating the de Broglie wavelength.

Planck's Constant

Planck's constant is a fundamental constant denoted by the symbol \( h \). This constant is pivotal in quantum mechanics, symbolizing the action of energy at a microscopic level. In value, it is approximately \( 6.63 \, \times \, 10^{-34} \mathrm{~Js} \). Planck’s constant helps bridge the understanding between wave and particle properties.

You can think of Planck's constant as a small measuring stick in the world of physics that relates the energy of photons to their frequency. It is a vital component in equations, including the de Broglie wavelength formula. Without it, relating wave characteristics to particle properties wouldn't be mathematically possible.
In the de Broglie wavelength formula, using Planck's constant allows physicists and students alike to calculate the wavelength of particles, no matter how small they are, such as protons traveling at high speeds.

Wavelength Calculation

Calculating the wavelength associated with any tiny particle, such as a proton, is done using the de Broglie wavelength formula. This formula is given by:

• \( \lambda = \frac{h}{mv} \)

Here, \( \lambda \) is the wavelength, \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.
Step-by-step, you need to substitute the known variables into this formula. For example, in the exercise calculation, with \( h = 6.63 \times 10^{-34} \mathrm{~Js} \), \( m = 1.67 \times 10^{-27} \mathrm{~kg} \), and \( v = 1.0 \times 10^{3} \mathrm{~ms}^{-1} \), you substitute these values into the formula.
Then you calculate it to find the de Broglie wavelength:

• \( \lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.0 \times 10^{3}} = 3.97 \times 10^{-10} \mathrm{~m} \)

Finally, to make the wavelength more understandable and often easier to work with, convert it into nanometers (nm), knowing that 1 nm = \( 10^{-9} \) meters. This makes managing these minuscule measurements more convenient, especially in scientific applications and exercises.