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प्रश्न
A cobalt specimen emits induced radiation of 75.6 millicurie per second. Convert this disintegration in to becquerel (one curie = 3.7 × 1010 Bq)
उत्तर
Cobalt specimen emits induced radiation = 75.6 millicurie per second
(1 curie = 3.7 × 1010 Bq)
So 75.6 millicurie = 75.6 × 103 × 1 curie
= 75.6 × 10-3 × 3.7 × 1010 Bq
= 279.72 × 107
= 2.7972 × 109 Bq
75.6 millicurie per second is equivalent to 2.7972 × 109 Bq.
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संबंधित प्रश्न
Man-made radioactivity is also known as _______.
Artificial radioactivity was discovered by ______.
In which of the following, no change in mass number of the daughter nuclei takes place
- α decay
- β decay
- γ decay
- neutron decay
Gamma radiations are dangerous because
________ aprons are used to protect us from gamma radiations.
_______ is used to measure exposure rate of radiation in humans.
The radio isotope of ________ helps to increase the productivity of crops.
Match: III
| a. | Soddy Fajan | Natural radioactivity |
| b. | Irene Curie | Displacement law |
| c. | Henry Bequerel | Mass energy equivalence |
| d. | Albert Einstein | Artificial Radioactivity |
Mark the correct choice as
- Assertion: In a β - decay, the neutron number decreases by one.
- Reason: In β - decay atomic number increases by one.
Who discovered natural radioactivity?
