Advertisements
Advertisements
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)
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
Man-made radioactivity is also known as _______.
In which of the following, no change in mass number of the daughter nuclei takes place
- α decay
- β decay
- γ decay
- neutron decay
Positron is an ________.
The radio isotope of ________ helps to increase the productivity of crops.
If the radiation exposure is 100 R, it may cause ______.
Match: III
| a. | Soddy Fajan | Natural radioactivity |
| b. | Irene Curie | Displacement law |
| c. | Henry Bequerel | Mass energy equivalence |
| d. | Albert Einstein | Artificial Radioactivity |
Who discovered natural radioactivity?
Give the SI unit of radioactivity.
Which material protects us from radiation?
Mr. Ramu is working as an X-ray technician in a hospital. But, he does not wear the lead aprons. What suggestion will you give to Mr. Ramu?
