diff --git a/.obsidian/workspace.json b/.obsidian/workspace.json
index d69ceca3..c046d44f 100755
--- a/.obsidian/workspace.json
+++ b/.obsidian/workspace.json
@@ -6,7 +6,7 @@
       {
         "id": "26fa757f3b505503",
         "type": "tabs",
-        "dimension": 46.035242290748904,
+        "dimension": 43.245227606461086,
         "children": [
           {
             "id": "7220717c2848cdda",
@@ -25,7 +25,7 @@
       {
         "id": "f2710f80feb0fcd8",
         "type": "tabs",
-        "dimension": 53.964757709251096,
+        "dimension": 56.75477239353891,
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diff --git a/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md b/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md
index 6d0e8b0a..f6a4bb28 100644
--- a/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md	
+++ b/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md	
@@ -50,16 +50,21 @@ $\left(8.98755 \times 10^{13} \text{ J } \right) \times \frac{1 \text{ kg }}{3 \
 	*a. To what nucleus does $^3\text{H}$ decay?*
 	[Helium-3](https://people.physics.anu.edu.au/~ecs103/chart/)
 	*b. What is the mass in grams of 1 mCi of tritium?*
-	$1 \text{ mCi } \times \left( 3.7\times 10^{10} \frac{\text{decay}}{\text{s}}\right)$
+	First, we need to find the decay constant of tritium:
+	$\lambda = \frac{0.693 \text{ decay}}{12.26 \text{ years}} = 1.79241 \times 10^{-9} \frac{\text{decay}}{\text{s}}$ 
+	And we also know that one millicurie is:
+	$1 \text{ mCi} = 3.7 \times 10^{10} \frac{\text{decay}}{\text{s}}$
+	Therefore we find multiple of the decay constant we need:
+	$$\text{Ratio} = \frac{1 \text{ mCi}}{\lambda} = \frac{3.7 \times 10^{10} \frac{\text{decay}}{\text{s}}}{1.79241 \times 10^{-9} \frac{\text{decay}}{\text{s}}} = 2.06426 \times 10^{19}$$
+	Then we know we need this many atoms to decay (on average) at the mean activity. We now can convert to grams:
+	$$\left(2.06426 \times 10^{19}\right) \times \frac{1 \text{ mol }^3H}{0.6022045 \times 10^{24} \text{atoms}} \times \frac{3.01605 \text{ g}}{1 \text{ mol } ^3H} = 1.03385 \times 10^{-4} \text{ g}$$
 	
-	
-
 **8. Approximately what mass of $^{90}\text{Sr}$ (T-1/2 = 28.8 years) has the same activity as 1g of $^{60}\text{Co}$ (T-1/2 = 5.26 years)?**
+First let's find the number of cobalt atoms:
+$$1 \text{g} \times \frac{1 \text{ mol } ^{60}\text{Co}}{59.934 \text{ g}} =  1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co}$$
+Now we can find how much more strontium we need:
+$$\frac{28.8 \text{ years}}{5.26 \text{ years}} = 5.47528$$
+Finally we multiply this number by the moles of cobalt, and convert back to mass for strontium-90:
+$$5.47528 \times \frac{1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co}}{}$$
+
 **9. Using the chart of the nuclides, complete the following reactions. If a daughter nucleus is radioactive, indicate the complete decay chain:**
-
-	
-
-
-
-
-