Question

Calculate the momentum and kinetic energy of a neutron with a wavelength of \(1.8 \times 10^{-10}\,\mathrm{m}\) and a mass of \(1.67 \times 10^{-27}\,\mathrm{kg}\).
• \(4.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}},\ 3.0 \times 10^{-21}\,\mathrm{J}\)
• \(5.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}},\ 6.0 \times 10^{-21}\,\mathrm{J}\)
• \(3.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}},\ 4.0 \times 10^{-21}\,\mathrm{J}\)
• \(7.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}},\ 8.0 \times 10^{-21}\,\mathrm{J}\)

• A. \(4.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}},\ 3.0 \times 10^{-21}\,\mathrm{J}\)
• B. \(5.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}},\ 6.0 \times 10^{-21}\,\mathrm{J}\)
• C. \(3.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}},\ 4.0 \times 10^{-21}\,\mathrm{J}\)
• D. \(7.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}},\ 8.0 \times 10^{-21}\,\mathrm{J}\)

Hint:

For microscopic particles: \[ p = \frac{h}{\lambda} \] Then use: \[ K.E. = \frac{p^2}{2m} \] Always handle powers of ten carefully while simplifying.

Answer 1

The Correct Option is A

Concept: According to de Broglie's hypothesis, every moving particle possesses wave nature. The de Broglie wavelength relation is: :contentReference[oaicite:0]{index=0} where:
• \( \lambda \) = wavelength
• \( h \) = Planck's constant
• \( p \) = momentum Momentum can therefore be written as: \[ p = \frac{h}{\lambda} \] The kinetic energy of a particle is: :contentReference[oaicite:1]{index=1} where:
• \(p\) = momentum
• \(m\) = mass of the particle

Step 1: Write the given values. Given: \[ \lambda = 1.8 \times 10^{-10}\,\mathrm{m} \] \[ m = 1.67 \times 10^{-27}\,\mathrm{kg} \] Planck's constant: \[ h = 6.626 \times 10^{-34}\,\mathrm{Js} \]

Step 2: Calculate the momentum using de Broglie relation. Using: \[ p = \frac{h}{\lambda} \] Substituting values: \[ p = \frac{6.626 \times 10^{-34}}{1.8 \times 10^{-10}} \] Now divide the numerical values: \[ \frac{6.626}{1.8} \approx 3.68 \] Apply laws of exponents: \[ 10^{-34} \div 10^{-10} = 10^{-24} \] Thus: \[ p \approx 3.68 \times 10^{-24}\,\mathrm{kg\,ms^{-1}} \] Rounding: \[ p \approx 3.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}} \]

Step 3: Calculate kinetic energy. Using: \[ K.E. = \frac{p^2}{2m} \] Substitute: \[ K.E. = \frac{(3.68 \times 10^{-24})^2}{2 \times 1.67 \times 10^{-27}} \] First square the momentum: \[ (3.68)^2 \approx 13.54 \] \[ (10^{-24})^2 = 10^{-48} \] Thus: \[ p^2 \approx 13.54 \times 10^{-48} \] Denominator: \[ 2m = 3.34 \times 10^{-27} \] Now divide: \[ K.E. \approx \frac{13.54 \times 10^{-48}}{3.34 \times 10^{-27}} \] \[ K.E. \approx 4.05 \times 10^{-21}\,\mathrm{J} \] Therefore: \[ K.E. \approx 4.0 \times 10^{-21}\,\mathrm{J} \]

Step 4: Match with the given options. We obtained: \[ p \approx 3.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}} \] \[ K.E. \approx 4.0 \times 10^{-21}\,\mathrm{J} \] This matches Option (3). Final Conclusion: The momentum and kinetic energy of the neutron are: \[ \boxed{ 3.7 \times 10^{-24}\,\mathrm{kg\,ms^{-1}} } \] and \[ \boxed{ 4.0 \times 10^{-21}\,\mathrm{J} } \] Hence, the correct answer is: \[ \boxed{(3)} \]