Question

I do't have a clue on solving this-Using inverse square law of radiation, what geiger counter readingwould you get for a radioactive material when you are 3 feet from the source(assuming an initial reading of 6300R AT A DISTANCE OF 1 FOOT from the source)

Answer 1

r is the radius or distance between the radioactive material and the counter

Answer 2

\[I(r)\propto\frac 1{r^2}\]\[I(r)=\frac k{r^2}\]\[r^2\times I(r)=k\]\[1[\text{ft}]^2\times6300\left[\text R\right]=k\]
\[I(3)=\frac{k}{\left(3[\text{ft}]\right)^2}\]

Answer 3

\[\Psi(x,t) = Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}\]\[\langle p^2 \rangle =?\]

\[A=\sqrt[4]{\frac{2am}{\pi\hbar}}\]\[\frac{\partial^2\Psi}{\partial x^2} =\left(\frac{-2ma}{\hbar}+ \frac{4m^2a^2x^2}{\hbar^2}\right)\Psi\]