Question

question 6 (1 point)
note that this is a follow up to question 5
210 206 b 0
rn → po + c + d
86 a 2 0
what two types of radiation is emitted in the above problem?
1 and 2
give answers in the order from left to right, ie a before b
a. radiation b. alpha c. beta d. gamma
e. neutron f. neutrino g. proton

Explanation:

Step1: Analyze mass - number change

The mass - number of Rn is 210 and of Po is 206. The decrease in mass - number is 210 - 206=4. An alpha particle (\(\alpha\)) has a mass - number of 4. So one of the emitted particles is an alpha particle.

Step2: Analyze atomic - number change

The atomic - number of Rn is 86. Let the atomic - number of Po be \(A\). Since an alpha particle has an atomic - number of 2, and considering the conservation of atomic - number in radioactive decay. If we assume the emission of an alpha particle first, then for the remaining decay, we note that the sum of atomic - numbers on the right side must equal 86. The atomic - number of Po is 84 (because \(86-2 = 84\)). The particle with atomic - number 2 is an alpha particle and the particle with mass - number 0 and atomic - number 0 is a gamma ray. Gamma rays (\(\gamma\)) are often emitted during radioactive decay to release excess energy.

Answer:

B. alpha, D. gamma