Question

The momentum of a particle having a de-Broglie wavelength of $10^{17}$ m is (Given, $h = 6.625 \times 10^{-34}$ m)

• A. $3.3125 \times 10^{-7} kg m s^{-1}$
• B. $26.5 \times 10^{-7} kg m s^{-1}$
• C. $6.625 \times 10^{-17} kg m s^{-1}$
• D. $13.25 \times 10^{-17} kg m s^{-1}$

Answer 1

The Correct Option is C

To find the momentum of a particle given its de-Broglie wavelength, we can use the de-Broglie wavelength formula:

\(\lambda = \frac{h}{p}\)

where:

• \(\lambda\) is the de-Broglie wavelength
• \(h\) is the Planck's constant, given as \(6.625 \times 10^{-34}\) m² kg / s
• \(p\) is the momentum of the particle

We need to rearrange this formula to solve for momentum (\(p\)):

\(p = \frac{h}{\lambda}\)

Given that the de-Broglie wavelength (\(\lambda\)) of the particle is \(10^{17}\) m, we substitute the given values into the formula:

\(p = \frac{6.625 \times 10^{-34} \, \text{m}^2 \, \text{kg/s}}{10^{17} \, \text{m}}\)

Calculate the momentum:

\(p = 6.625 \times 10^{-34} \cdot 10^{-17} \, \text{kg m/s}\)

\(p = 6.625 \times 10^{-51} \times 10^{17} \, \text{kg m/s}\)

\(p = 6.625 \times 10^{-34} \, \text{kg m/s}\)

The calculated value of the momentum matches with one of the given options:

\(6.625 \times 10^{-17} \, \text{kg m/s}\)

Hence, the correct answer is:

\(6.625 \times 10^{-17} \, \text{kg m/s}\)

This concludes the step-by-step solution to find the momentum of the particle.