Question

The wavelength of an electron of kinetic energy 4.50 × 10-29 J is...... × 10-5 m. (Nearest integer)

Answer 1

7

Answer 2

The de Broglie wavelength ($\lambda$) of a particle is given by the formula $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle. Momentum can be related to kinetic energy (KE) using the formula $p = \sqrt{2mKE}$, where $m$ is the mass of the particle. Substituting this into the de Broglie equation gives: $\lambda = \frac{h}{\sqrt{2mKE}}$ Given values are: Planck's constant, $h = 6.6 \times 10^{-34}$ J s Mass of electron, $m = 9 \times 10^{-31}$ kg Kinetic energy, $KE = 4.50 \times 10^{-29}$ J

Substitute these values into the formula: $\lambda = \frac{6.6 \times 10^{-34} \text{ J s}}{\sqrt{2 \times (9 \times 10^{-31} \text{ kg}) \times (4.50 \times 10^{-29} \text{ J})}}$ First, calculate the term inside the square root: $2mKE = 2 \times 9 \times 10^{-31} \times 4.50 \times 10^{-29} = 81 \times 10^{-60} \text{ kg}^2\text{m}^2/\text{s}^2$ Now, take the square root: $\sqrt{2mKE} = \sqrt{81 \times 10^{-60}} = 9 \times 10^{-30} \text{ kg m/s}$ Now, calculate the wavelength: $\lambda = \frac{6.6 \times 10^{-34} \text{ J s}}{9 \times 10^{-30} \text{ kg m/s}} = \frac{6.6}{9} \times 10^{-34 - (-30)} \text{ m}$ $\lambda = \frac{6.6}{9} \times 10^{-4} \text{ m}$ Calculating the fraction: $\frac{6.6}{9} \approx 0.7333...$ So, the wavelength is: $\lambda \approx 0.7333... \times 10^{-4} \text{ m}$ The question asks for the answer in the format ...... × 10-5 m. Convert the wavelength to this format: $\lambda \approx 0.7333... \times 10^{-4} \text{ m} = 7.333... \times 10^{-5} \text{ m}$ The question requires the nearest integer value for the blank space. The value is $7.333...$. The nearest integer to $7.333...$ is $7$.