Question 26 3 points if a radioactive isotope with an...

Question 26 (3 points): If a radioactive isotope with an initial amount of 1.00 gram has a measured half-life of only 4.00 hours, then in hours, how long will it take before its remaining mass is 0.0156 grams? Question 26 options: a) 6.00 hours b) 12.0 hours c) 18.0 hours d) 24.0 hours e) 30.0 hours Question 27 (3 points): Which type of nuclear reaction is mostly responsible for the tremendous energy output of our sun and all the stars in the universe? Question 27 options: a) Nuclear fusion b) Nuclear fission c) Beta-decay d) Alpha decay

Transcript

00:01 Hello students, for the question number 38, so the radioactive material has a half life of one day, so t half is equal to one day and the initial amount of material present that is n naught is equal to 1024 grams and we have to calculate the material that is present at the end of, so first one is equal to three days and we have to calculate the material that is present at the end of six days and the material that remains after a time of 10 days, so these three things we have to calculate that is the material that is remaining n we have to calculate, so this is calculated by the formula n equal to n naught into 1 by 2 t by t half, so this is the formula we are using here to calculate this, so we are going to calculate them simultaneously, so here for t is equal to three days n is equal to n naught is 1024 into 1 by 2, t here is three days divided by t half is one day, this will give 1024 divided by 1 cube 1 2 cube 8, so 1024 by 8, so the remaining material is 128 grams at the end of three days 128 grams will remain, similarly for the second one that is 1224 into 1 by 2 into 10 by 1, so 1024 divided by 2 power 10 is 1024, so this will give n is equal to 1 gram, so these are the amount of substance that will remain after 3 days 6 days and 10 days respectively and now we are moving over to the second question that is question number 39, so for question number 39 that t half is equal to 30 years and we have to calculate the time required for the substance to reduce to one sixth of the initial value, so let initial value or initial amount is equal to n naught and the final amount is so final amount is n is equal to 1 by 6 n naught, so this can be calculated from the formula n is equal to n naught into 1 by 2 t by t half, so here n is equal to 1 by 6 n naught equal to n naught into 1 by 2, so here t we want to calculate and by t half is 30, so n naught and n naught will get cancelled, so 1 by 6 is equal to 1 by 2 into t by 30, so if we solve this equation we get the time as 77 .54 years, so the time required for one sixth of the material to disintegrate or decay is equal to, so time required for one sixth of the material to decay is equal to 77 .54 years...

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