Dimensional Analysis to Determine a Formula The kinetic energy of a particle confined to a spherical region with a uniform
internal potential depends on its mass, the radius of the sphere, and the Planck constant. An electron, confined to such a region of radius $1$ nm, has kinetic energy $0.38$ eV. Use dimensional analysis to find the kinetic energy of a proton confined to a region the size of an atomic nucleus (radius ≈ $5 × 10^{−15}$ m).
I cannot produce a formula which results in $0.38$ eV as in the example.
 A: We use the notation $[X] = L^{a} T^{b} M^{c}$ to denote the dimension of the quantity $X$, which is of length to the power $a$, time to the power $b$ and mass to the power $c$. In your problem, you are supplied with


*

*Mass, $[m] = M$

*Radius of the sphere, $[r]=L$

*Planck's constant $[h] = L^2 M T^{-1}$


We seek a quantity, potential energy, with dimensions $[E] = L^2MT^{-2}$. We demand that a combination $m^{x}r^yh^z$ has the same dimensions, providing $x,y$ and $z$ are suitably chosen. This leads to a set of three simultaneous linear equations, i.e.
$$x +z = 1$$
$$y + 2z = 2$$
$$-z = -2$$
It is immediately clear, $z = 2$. Therefore, $y=-2$ from the second equation, and $x=-1$ from the first equation. We therefore obtain a formula for the energy, 
$$E = \frac{1}{m}\left(\frac{h}{r}\right)^2$$
which is dimensionally correct. The actual expression for the energy of the particle differs at most by a dimensionless constant, providing the variables supplied are the only relevant dimensionful quantities.
A: The constant for both problems, that of the proton and that of the electron, is the same, and it can calculated from the data for the electron, as shown below.
VARIABLES ....................              DIMENSIONS
Kinetic energy ke ....................                  $ML^2T^{–2}$
Mass m ...................................                      $M$
Radius r ..................................                     $L$
Planck’s constant h ................                $ML^2T^{–1}$
    Planck’s constant  6,62607 x 10^–34
    1 eV = 1,6020 x 10^–19 J
    Mass of the electron  0,910904 x 10^–30 kg
    Mass of the proton   1,6726219 × 10^–27 kg

The variables form a non-dimensional product k
$k = ke^a m^b r^c h^d$  where a,b,c,d are numbers to be determined.
Let’s form now a parallel product k* with the dimensions:
$k* = (ML^2T^{–2})^a  (M)^b  (L)^c (ML^2T^{–1})^d$
Clearly, $k* = L^0M^0T^0$. We now take the exponents for each dimension:
a  + b + d = 0
2a + c + 2d = 0
–2a – d = 0
We make a = 1, since ke is the variable we’re going to solve for. 
1  + b + d = 0
2 + c + 2d = 0
–2 – d = 0
d = –2 .... b = 1 .... c = 2
Then, 
$k = ke^a m^b r^c h^d$  =>      $k = ke^1 m^1 r^2 h^{–2}$ 
We know the all the data for the electron, and we are interested in obtaining the constant, that is valid for the general problem, and so we’ll use it for the calculation of the case of the proton. Hence, we insert values, taking care to use the same units.
$k = 6,0876 \times 10^{–20} · 0,910904 \times 10^{–30} · 1 \times 10^{–18} · 2,28135 \times 10^{66}$
k = 0,126506
We have now k. Solving the expression above for ke, we have:
$ke = k  m^{–1} r^{–2} h^2$ 
Replacing values for the case of the proton:
$ke = 0,126506 · 5,9786 \times 10^{26} · 4 x 10^{28} · 4,39048 x 10^{–67}$
$ke = 1,328258 \times 10^{–12} J  = 8,29125 \times 10^6 eV$
