/sci/ - Science & Math

[Anonymous No.4714941]

How much bananas do I need to eat to become dangerously radioactive?

[Anonymous No.4714945]

Bump

[Anonymous No.4714947]

OVER 9000!

[Anonymous No.4714949]

four

[Anonymous No.4714950]

>>4714947
i want some math.

[Anonymous No.4714954]

You would die from eating too much rather than radioactivity.

[Anonymous No.4714958]

>>4714954
i didnt say how many i would need to eat to die. I want to know how many i need to become dangerously redioactive.

[Anonymous No.4714979]

Followup: How long can I spend in Fukushima before I become dangerously radioactive?

[Anonymous No.4714985]

OP, Just eat the Americium in smoke alarms. It's radioactive and actually possible to eat enough to be dangerous

[Anonymous No.4714988]

>>4714941

You need to eat about 30 before your feces will burn right through your intestinal wall. From there, it will just burn through the skin and fall onto the floor. Be careful, OP, that's how my friend's grandfather died. His radioactive feces caused his colostomy bag to catch fire.

[Anonymous No.4714996]

>>4714985
okay, here are the variables i need.

How many beta (or positron i honestly dont know how potassium decays) particles are needed to become dangerous.

And

How many particles are given off by potassium, lets assume it's not past it's first halflife.

[Anonymous No.4715002]

A Banana is 0.1 µSv
Now do the math

[Anonymous No.4715015]

>>4715002
so ill use the yearly max or public, 1mSU. Well math makes that 10000 bananas.... Great, ill get started.

[Math300K !BeGauss9DU No.4715063]

1 gram of potassium has 0.000117 g of K40
The half life of the radioactive isotope of potassium( K40 ) is 1.25 billion years (4×10^16 seconds) therefore the activity is of 31 Bq/g (in other words: 1 gram of potassium releases 31 atoms each second)
EPA says the conversion factor is 5.02 nanosieverts over 50 years for each becquerel of isotopically pure potassium-40 ingested by an average adult.
Wolfram alpha says that a banana contains 1.5 grams of K.
So we get:
5.02 nSv/Bq × 31 Bq/g × 1.5 g =233 nSv = 0.233 μSv.
We're gonna be optimistic so we use the fact that a single dose of 5,000 mSv which would kill about half of those receiving it within a month.
5000/0.233 = 22,421 BANANAS
The average banana weights 118 grams(Wolf-alpha) so you'd have to EAT 2645.74 tons of this lovely fruit.

[Anonymous No.4715074]

>>4715063
>dividing m by u
Multiply by 1000 OP

[Math300K !BeGauss9DU No.4715078]

>>4715063
sorry I dun goofed

it's:
5000000/0.233 = 21,459,227 bananas
2,645,740 Tons

[Anonymous No.4715082]

>>4715078
thats a ton of bananas!

[Anonymous No.4715091]

>>4715078
wtf, 2645740 tonns/ 21459227 bananas = .123 tonns per banana

So wtf, bananas arent that heavy.

[Math300K !BeGauss9DU No.4715095]

>>4715074
I noticed when I read my post again
I was like...wut..being surrounded by only 2500 tons would kill you?

>>4715082
Banana aren't dangerous... our body contain potassium as well: about 2.5 g per Kg:
75 Kg × 2.5g/Kg × 31 Bq/g = 5400 Bq of radiation

Our body has the same potassium as 125 bananas, so there's no reason to worry.

[Math300K !BeGauss9DU No.4715101]

>>4715091
>>4715091
Sorry bout that. I'm pretty stupid at 2 AM
It's Kg...divide by 1000 again

[Anonymous No.4715107]

>>4715101
haha its ok man, you helped anyway.