A particle of mass 6.63 \times 10^{-27}\text{ kg} has a de Broglie wavelength of 2.0 \time… — JAMB UTME

A particle of mass $6.63 \times 10^{-27}\text{ kg}$ has a de Broglie wavelength of $2.0 \times 10^{-13}\text{ m}$ . What is the speed of the particle in $\text{m/s}$ ? (Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ )

Answer: 500000 m/s

Answer

The speed of the particle is

500,000\text{ m/s}

(or

5.0 \times 10^5\text{ m/s}

According to de Broglie's hypothesis, the matter wavelength

\lambda

of a particle is related to its momentum

p = m v

\lambda = \frac{h}{m v}

. Rearranging for speed yields

v = \frac{h}{m \lambda}

. Substituting

h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}

m = 6.63 \times 10^{-27}\text{ kg}

, and

\lambda = 2.0 \times 10^{-13}\text{ m}

gives

v = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-27} \times 2.0 \times 10^{-13}} = 500,000\text{ m/s}

Step-by-Step Solution

Identify the given physical quantities and formula

m = 6.63 \times 10^{-27}\text{ kg}

\lambda = 2.0 \times 10^{-13}\text{ m}

h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}

. The de Broglie equation is

\lambda = \frac{h}{m v}

The de Broglie wavelength formula relates matter wave properties to particle momentum.

Isolate the target variable (speed

v

)

v = \frac{h}{m \lambda}

Algebraic manipulation is needed to solve directly for speed.

Substitute parameters and compute the result

v = \frac{6.63 \times 10^{-34}}{(6.63 \times 10^{-27})(2.0 \times 10^{-13})} = \frac{6.63 \times 10^{-34}}{1.326 \times 10^{-39}} = 500,000\text{ m/s}

Performing scientific notation division yields the numerical particle speed.

Key Concept

de Broglie Wavelength and Particle Speed

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