Question

if you have a sample of U- 235 with a mass of 50g and it has undergone 4 half-lives. how much did the original sample weigh ? please show step by step and the equation thanks ...

Answer 1

After undergoing one **half life, how much mass is there if the original sample had mass M?

Answer 2

if you have a sample of U- 235 with a mass of 50g and it has undergone 4 half-lives. how much did the original sample weigh .. this is the only information the question is asking

Answer 3

Right, but do you know what "half life" means?

Answer 4

no

Answer 5

can u help me solved this please i need to finished for homework

Answer 6

^_^ If something has undergone 1 half life, then it had 1/2 (half) its original mass.

\[m_1=\frac{1}{2}M\]

If something has undergone 2 half lives, then it has half of half (1/2)(1/2) its original mass

\[ m_2 = \frac{1}{2} m_1 = \frac{1}{2} \frac{1}{2} M = \frac{1}{4}M\]

So what would it look like if it has undergone 4 half lives?

Answer 7

sorry i dont get it at all

Answer 8

my teacher give me a equation Mo(1/2)^n=MR

Answer 9

mo= original mass ..n-#of half lives passed mr-remaining mass

Answer 10

Right, exactly

Answer 11

so if MR = 50g, and n=4, solve for Mo

Answer 12

so x=(1/2)^n=50 g ?

Answer 13

\[ \frac{M_0}{16} = M_R\]
\[M_0 = 16 M_R\]

Answer 14

\[ M_0 = 16*50 g\]

Answer 15

i need an equation please

Answer 16

\[ M_0\Big( \frac{1}{2} \Big) ^4 = M_R\]

\[M_0 \frac{1}{16} = M_R\]
\[M_0 = 16 M_R\]
\[M_0 = 16 ( 50g)\]
\[M_0 = 800 g\]

Answer 17

thanks can u help me with anohter one

Answer 18

Sure - the question asks for the weight of the sample though, and weight is mass times gravity.

Answer 19

Thorium-234 has a half life of 24 days. If 360 days pass and you have only 0.5g of the originalsample, what was the mass of the original sample ?

Answer 20

can u identify MR , N and mo for me please

Answer 21

So what's MR in the question? (there's only one mass)

Answer 22

is ask for the mass of the original sample

Answer 23

Right, I get that.

Answer 24

I'm asking *you

Answer 25

y it say original is 0.5 bit it ask for original sample

Answer 26

this is y im so confusing

Answer 27

MR = 0.5 g

It's 0.5 g after 360 days

Half life is 24 days.

360 / 24 = ?

Answer 28

360/24=15

Answer 29

so N=15

Answer 30

MR = 0.5g
N = 15

Answer 31

so is x(1/2)=0.5

Answer 32

M0 (1/2)^15 = MR

M0 (1/2)^15 = 0.5g

M0 = 2^15 (0.5g)

Answer 33

my teacher give usthe answer is 2x 10 ^4

Answer 34

the the final answer is not correct compare to her did we do something wrong ?

Answer 35

she gimy teacher giveme the answer but no work so i dont know how
she get that

Answer 36

2^15 = 32768

32768 * (.5g) = 16384g

You only have 1 significant figure in the problem so

16384g rounds up to 20000g

20000g = 2 x 10^4g

Answer 37

where u get 2?

Answer 38

i thought .5^15 where u get 2 ^15 ?

Answer 39

\[ M_0 \Big(\frac{1}{2} \Big)^{15} = M_R\]
\[ M_0 \frac{1}{2^{15}} = M_R\]
\[ M_0 = 2^{15}M_R\]

Answer 40

yea but when i put in the equation how i plot it in ?

Answer 41

?

Answer 42

mo=(2)^15=0.5 ?

Answer 43

\[M_0 = 2^{15}(0.5g)\]

Answer 44

ok thanks

Answer 45

r u a teacher u are good

Answer 46

Just a tutor, but thanks! ^_^

Answer 47

how can i contact u next time if i need help

Answer 48

You can message, fan, or just tag me in your next question - hopefully the site will be finished with lots of its updates soon ^_^