Luis Alvarez's K-T Impactor Calculation
The volume of a sphere is $\small\sf{\frac4 3\pi R^3}$, where $\small\sf{R}$ is the radius.

The volume of Earth with the 4 cm deep iridium rich layer is,

$\small\sf{V_{Ei} = \frac 4 3\pi (6.378\cdot10^6)^3}$ $\small\sf{m^3}$ = $\small\sf{1.086 \ 781 \cdot 10^{21}}$ $\small\sf{m^3}$

The volume of the 4 cm deep iridium rich layer is,

$\small\sf{V_i = \frac 4 3\pi (6.378\cdot10^6)^3} -\frac 4 3\pi (6.378\cdot10^6 - 0.04)^3 $ $\small\sf{m^3}$ = $\small\sf{2.044 \cdot 10^{13}}$ $\small\sf{m^3}$

Now, density is mass divided by volume, $\small\sf{\rho = m/v}$, thus the mass of the iridium rich layer is,

$\small\sf{m_i = 2.5 \cdot 2.044 \cdot 10^{13} = 5.110 \cdot 10^{13}}$ tonnes

The proportion of iridium in the layer is 10 ppb, therefore the mass of iridium within the layer is,

$\small\sf{5.110 \cdot 10^{13} \cdot 10 \cdot 10^{-9}} = 511 \ 000$ tonnes

Now, the proportion of iridium in the meteor was 0.5 ppm, therefore the mass of the meteor was,

$\small\sf{511 \ 000 / (0.5 \cdot 10^{-6}) = 1.0220 \cdot 10^{12}}$ tonnes

The density of the meteor was  $\small\sf{6.0 \ g/cm^3}$, which is the same as $\small\sf{6.0 \ t/m^3}$, therefore, the volume of the meteor was,
$\small\sf{1.0220 \cdot 10^{12}/6.0} = 1.703 \ 333 \cdot 10^{11} m^3$

Using this and the equation for the volume of a sphere, the radius of the meteor was,

$\small\sf{\sqrt[3](\frac 3{4\pi} \cdot 1.703 \ 333 \cdot 10^{11})} = 3438.774 \ m$

and the diameter of the meteor was 6877.5 m or 6.878 km

Not the 10 km you wanted.



To get the thickness you want for a 10 km diameter meteor, do the calculation in reverse.

For diameter of 10 000 m, the volume of a spherical meteor would have been,

$\small\sf{\frac 4 3 \cdot \pi \cdot 5000^3} = 523.598 \ 776 \cdot 10^9 \ m^3$

With a density of $\small\sf{6 \ t/m^3}$, the mass of meteor would have been,

$\small\sf{6(523.598 \ 776 \cdot 10^9)} = 3.141 \ 592 \ 6 \cdot 10^{12} t$

with a metal grade of 0.5 ppm of iridium, the mass of iridium in the meteor would have been,

$\small\sf{3.141 \ 592 \ 6 \cdot 10^{12} \cdot 0.5 \cdot 10^{-6} = 1 \ 570 \ 796 \ t}$

The metal grade of iridium in the layer on the Earth is 10 ppb, therefore the mass of the iridium layer is,

$\small\sf{(1 \ 570 \ 796)/ (10 \cdot 10^{-9}) = 1.570 \ 796\ 327 \cdot 10 ^{14} \ t}$

With a density of $\small\sf{2.5\  t/m^3}$, the volume of the layer is,
$\small\sf{(1.570 \ 796\ 327 \cdot 10 ^{14})/2.5 = 6.283 \ 185 \cdot 10^{13} \ m^3}$

From the previous calculation, the volume of layer is,

$\small\sf{6.283 \ 185 \cdot 10^{13} = \frac 4 3\pi (6.378\cdot10^6)^3} -\frac 4 3\pi (6.378\cdot10^6 - x)^3 $ $\small\sf{m^3}$ = $\small\sf{2.044 \cdot 10^{13}}$ $\small\sf{m^3}$

Solving for $\small\sf{x}$, which is the thickness of the layer, gives $\small\sf{x} = 0.123\ m\ or\ 12.3\ cm$