If 2 α particles and 3 β particles emit in series decay of a radioactive nucleus Ra 88 228 and nucleus Y is obtained at last then find atomic number and mass number of Y .

In first, the radioactive Ra 228 88 nucleus. 2 α particles and 3 β particles emit in series decay of a radioactive nucleus. Then in 2 α decay, Ra 228 88 → - 2 α Z 220 84 In 3 β decay, Z 220 84 → - 3 β Y 220 87 Atomic number of Y = 87 Atomic number of A = 220 Ra 228 88 → Z 220 84 + 2 He 4 2 α particle Z 220 84 → Y 220 87 + 3 β ° - 1 β particle Hence, it is the required answer.

If 2 α particles and 3 β particles emit in series decay of a radioactive nucleus Ra 88 228 and nucleus Y is obtained at last then find atomic number and mass number of Y .

In first, the radioactive Ra 228 88 nucleus. 2 α particles and 3 β particles emit in series decay of a radioactive nucleus. Then in 2 α decay, Ra 228 88 → - 2 α Z 220 84 In 3 β decay, Z 220 84 → - 3 β Y 220 87 Atomic number of Y = 87 Atomic number of A = 220 Ra 228 88 → Z 220 84 + 2 He 4 2 α particle Z 220 84 → Y 220 87 + 3 β ° - 1 β particle Hence, it is the required answer.

A radioactive material reduces to 1 16 th of its initial value in 500 years. Find the half life of radioactive material.

R R 0 = ( 1 2 ) n 1 16 = ( 1 2 ) n ⇒ ( 1 2 ) 4 = ( 1 2 ) n ⇒ n = 4 T 1 / 2 = t n = 500 4 = 125 y e a r So half life of radioactive material is 125 years.

A radioactive material reduces to 1 16 th of its initial value in 500 years. Find the half life of radioactive material.

R R 0 = ( 1 2 ) n 1 16 = ( 1 2 ) n ⇒ ( 1 2 ) 4 = ( 1 2 ) n ⇒ n = 4 T 1 / 2 = t n = 500 4 = 125 y e a r So half life of radioactive material is 125 years.

What is meant by half-life of a radioactive material? Half-life of a radioactive material is 60 year. After how much time it becomes 1 16 of its initial value?

Half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive. R = R 0 / 16 T 1 2 = 60 U sin g . 693 T 1 2 t = ln R 0 R ∴ 0 . 693 60 × t = ln ( 16 ) ⟹ t = 60 0 . 693 × 2 . 77258872 = 239 . 82684 years