Question

20) Before a nuclear reaction, there is 0.0100 kg of uranium. After the nuclear reaction, there is 0.0090 kg of uranium. How much energy was released?

Answer 1

if someone can help me step by step ?

Answer 2

I believe you must use the formula \[E=∆ m{c}^{2}\] and here \(∆m=0.0100kg-0.0090kg\) and \(c=2.998 \times 10^8 m s^{-1}\)

Answer 3

ohh okay, yes I remember that equation! thank you . that makes sense

Answer 4

Yeah. You'll get units of \(kg~m^2~s^{-2}\) and that's the same as 1 J

Answer 5

wait it looks like the answer was :/
4.50 × 10^13 J

Answer 6

(h = 6.63 × 10^-34 J • s) they provided me with this for the question as well if that helps any

Answer 7

i still want to figure out how to solve it to get the 4.50 *10^13 j , just for my understanding. do you have any idea?

Answer 8

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jennylove
(h = 6.63 × 10^-34 J • s) they provided me with this for the question as well if that helps any
\(\color{#0cbb34}{\text{End of Quote}}\)
That's Planck's constant..so I'm assuming they solved this with Planck's formula \(E=hv\)

Answer 9

v is frequency...is there anything about frequency provided? Or perhaps wavelength?

Answer 10

no, just the question that was given above^ , thats about it. hmm thats confusing :/

Answer 11

Are you sure the answer is 4.50 × 10^13 J? I'm getting 8.928 * 10^13 J to 4 s.f.

Answer 12

thats what it says,unless theres an error:/ . how did you get 8.928 ?

Answer 13

oh wait no, im sorry yes its saying 9.0 × 10^13 J !

Answer 14

\[E=∆mc^2\]
\[E=(0.0100 kg - 0.0090 kg)(2.998 \times 10^8 m s^{-1})^2\]
\[8.928 \times 10^{13}~kg~m^2~s^{-2}=8.928 \times10^{13} J\] to 4 s.f.

Answer 15

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jennylove
oh wait no, im sorry yes its saying 9.0 × 10^13 J !
\(\color{#0cbb34}{\text{End of Quote}}\)
lol it's fine xD

Answer 16

oh okay so you just multiply it be plancks constant

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jennylove
oh okay so you just multiply it be plancks constant
\(\color{#0cbb34}{\text{End of Quote}}\)
Nah I didn't use Planck's constant, h, here. I'm using the constant for the speed of light. You would use Planck's constant in Planck's formula or some manipulation of it, which is \(E=hv\). The formula we used here to solve this problem is the well-known Einstein's formula

Answer 18

wait no , you did not use it. im sorry! i need to reread before I speak.

Answer 19

ohh okay , constant for speed! ill take note of that. let me reread everything and make sure i understand it. sorry about that

Answer 20

All g

Answer 21

just solved it myself, and now im going to take note of it. thank you for your help . I appreciate it