question 8\n6. given that e_kinetic = p^2/2m and the debroglie relationship given in the introduction…

question 8\n6. given that e_kinetic = p^2/2m and the debroglie relationship given in the introduction, calculate the constant that relates p with λ.\nfirst show your work to find the equation with the variables rearranged.\nconstant=\np = √(2m * e_kinetic)\np = √(2 * 9.11×10^(-31) kg * 3.2×10^(-22) j) = 2.41×10^(-26)\nconstant formula: λ * p\nfind constant: 2.8×10^(-8) * 2.41×10^(-26) = 6.75×10^(-34)\nquestion 9\nnow plug in your numerical values to find the value of the constant.\nconstant = 6.75×10^(-34) j×s

question 8\n6. given that e_kinetic = p^2/2m and the debroglie relationship given in the introduction, calculate the constant that relates p with λ.\nfirst show your work to find the equation with the variables rearranged.\nconstant=\np = √(2m * e_kinetic)\np = √(2 * 9.11×10^(-31) kg * 3.2×10^(-22) j) = 2.41×10^(-26)\nconstant formula: λ * p\nfind constant: 2.8×10^(-8) * 2.41×10^(-26) = 6.75×10^(-34)\nquestion 9\nnow plug in your numerical values to find the value of the constant.\nconstant = 6.75×10^(-34) j×s

Answer

Explanation:

Step1: Recall de - Broglie relation

The de - Broglie relation is $\lambda=\frac{h}{p}$, where $h$ is Planck's constant. Rearranging it to get the relation between $\lambda$ and $p$ in the form of a constant gives $h = \lambda\times p$.

Step2: Identify the constant

The constant that relates $p$ with $\lambda$ is Planck's constant $h$.

Answer:

$6.63\times 10^{-34}\ J\cdot s$ (Note: The correct value of Planck's constant is approximately $6.63\times 10^{-34}\ J\cdot s$, the value $6.75\times 10^{-34}$ in the provided work has some calculation error)