Question

Before a nuclear reaction, there is 0.01000 kg of uranium. After the nuclear reaction, there is 0.00950 kg of uranium. How much energy was released? (h = 6.63 × 1034 Js)

Answer 1

here we have to apply the Einstein's relationship between mass and energy, namely:
\[\Large \Delta E = \Delta m \cdot {c^2}\]
where:
\[\Large \Delta m = 0.01 - 0.0095 = ...{\text{Kg}}\]

Answer 2

5x10^-4?

Answer 3

\[\Large \Delta E = \Delta m \cdot {c^2} = 0.0045 \times 9 \times {10^{16}} = ...{\text{Joules}}\]

Answer 4

so 4.50x10^13 J?

Answer 5

that's right!

Answer 6

thanks so much! Could you help me with a couple more if you dont mind?

Answer 7

sorry I have made a typo the right formula is:
\[\Large \Delta E = \Delta m \cdot {c^2} = 0.0005 \times 9 \times {10^{16}} = 4.5 \times {10^{13}}\;{\text{Joules}}\]

Answer 8

ok! I will help you!

Answer 9

Thanks so much!
Which of the following processes is used by NASA to make electric power on deep space probes?

radioactive decay

fission

fusion

X-ray bombardment

Answer 10

I'm sorry I don't know that answer, since I'm Italian, and I never studied, at my university the technologies used by NASA

Answer 11

oh okay, could you help me with another one with some calculating values?

Answer 12

ok!

Answer 13

What is the energy of a single photon of red light with a wavelength of 700.0 nm?

Remember that E = hf and Planck's constant is expressed by h = 6.63 × 1034 Js = 4.14 × 1015 eVs.

2.84 × 1019 J

4.98 × 1019 J

2.84 × 1014 J

4.29 × 1014 J

Answer 14

we can write:
\[\lambda \nu = c\]
where \lambda is the wavelenght of your radiation \nu is its frequency, and c is the light speed
so we have:
\[\nu = \frac{c}{\lambda }\]

Answer 15

now the anergy of a photon, which is the mediator particle of the electromagnetic field, is given by the subsequent expression:
\[\Large E = h\nu = \frac{{hc}}{\lambda }\]

Answer 16

so you have to substitute your data into the above formula, what do you get?

Answer 17

Are we looking for c?

Answer 18

I got this:
\[\Large E = h\nu = \frac{{hc}}{\lambda } = 2.84 \times {10^{ - 19}}Joules\]

Answer 19

keep in mind that:
\[\Large h = 6.62 \times {10^{ - 34}}Joules \times \sec \]

Answer 20

so do we multiply the two?

Answer 21

here are more steps:
\[\Large E = h\nu = \frac{{hc}}{\lambda } = \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{700 \times {{10}^{ - 9}}}}\]

Answer 22

so 2.85x10^-37?

Answer 23

no, 2.84*10^(-19)

Answer 24

hmm, i dont know how I got that...

Answer 25

hint:
\[700nm = 700 \times {10^{ - 9}}meters = 7 \times {10^{ - 7}}meters\]

Answer 26

ohh okay, I put it in incorrectly!
do you have time for another?

Answer 27

yes!

Answer 28

A paleontologist measures the ratio of carbon-14 to carbon-12 in a fossil skull found at a site. What technique is he using and for what purpose?
(Points : 3)
using alpha emission to determine the structure of the skull

using radioactive decay to date the skull

using radioactive decay to determine the structure of the skull

using gamma emissions to date the skull

Answer 29

The ratio C14 to C12 is used to get the date of a sample which contains carbon, and the radioactive law is employed to get that date from that sample

Answer 30

more precisely, using the radioactive law, we can get when a sample has ceased to live

Answer 31

so, what is the right option?

Answer 32

using radioactive decay to date the skull?