what is the wavelength of a proton (m = 1.67×10^(-27) kg) moving at 1.2×10^5 m/s? (e_kinetic = ½ mv²; mass_e…

Explanation:

Step1: Identify the de - Broglie wavelength formula

The de - Broglie wavelength formula is $\lambda=\frac{h}{mv}$, where $h = 6.63\times10^{-34}\ J\cdot s$ (Planck's constant), $m$ is the mass of the particle, and $v$ is the velocity of the particle.

Step2: Substitute the given values

We are given that $m = 1.67\times10^{-27}\ kg$ and $v=1.2\times 10^{5}\ m/s$. Substitute these values into the formula: $\lambda=\frac{6.63\times 10^{-34}\ J\cdot s}{(1.67\times 10^{-27}\ kg)\times(1.2\times 10^{5}\ m/s)}$.

Step3: Calculate the wavelength

First, calculate the denominator: $(1.67\times 10^{-27}\ kg)\times(1.2\times 10^{5}\ m/s)=1.67\times1.2\times10^{-27 + 5}\ kg\cdot m/s=2.004\times10^{-22}\ kg\cdot m/s$. Then, $\lambda=\frac{6.63\times 10^{-34}\ J\cdot s}{2.004\times10^{-22}\ kg\cdot m/s}=3.31\times10^{-12}\ m$. Since $1\ nm = 10^{-9}\ m$, convert the wavelength to nanometers: $\lambda=3.31\times10^{-12}\ m\times\frac{1\ nm}{10^{-9}\ m}=0.00331\ nm\approx0.0033\ nm$.

Answer:

$0.033\ nm$