Calculate the wavelength (in nanometre) associated with a proton moving at \( 1.0 \times 10^3 \text{ m s}^{-1} \).

Question

Accepted Answer

Detailed Step-by-Step Solution

This problem is based on De Broglie's Equation, which relates the wavelength of a moving particle to its momentum.

The De Broglie wavelength (\(\lambda\)) is given by:

\lambda = \frac{h}{p} = \frac{h}{mv}

Planck's constant (\(h\)) = \( 6.63 \times 10^{-34} \text{ J s} \) (or \(\text{kg m}^2\text{s}^{-1}\))
Mass of proton (\(m\)) = \( 1.67 \times 10^{-27} \text{ kg} \)
Velocity (\(v\)) = \( 1.0 \times 10^3 \text{ m/s} \)

\lambda = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27}) \times (1.0 \times 10^3)}

First, simplify the denominator by combining the exponents of 10:

\text{Denominator} = 1.67 \times 10^{-27+3} = 1.67 \times 10^{-24} \text{ kg m/s}

Now, divide the numerator by the denominator:

\lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-24}} \) \( \lambda = \left( \frac{6.63}{1.67} \right) \times 10^{-34 - (-24)} \text{ m} \) \( \lambda \approx 3.97 \times 10^{-10} \text{ m}

Rounding off \(3.97\) to \(4.0\), we get:

\lambda \approx 4.0 \times 10^{-10} \text{ m}

We know that \( 1 \text{ nm} = 10^{-9} \text{ m} \). To convert meters to nanometres, multiply the result by \( 10^9 \):

\lambda \text{ (in nm)} = (4.0 \times 10^{-10}) \times 10^9 \text{ nm} \) \( \lambda = 4.0 \times 10^{-1} \text{ nm} \) \( \lambda = 0.40 \text{ nm}

Conclusion: The calculated De Broglie wavelength is 0.40 nm. Therefore, the correct option is (b).