| ⇦ | 
| ⇨ |
When ₉₀Th²²⁸ transforms to ₈₃Bi²¹², the number of emitted α and β particles are respectively
Options
(a) 8 α,7 β
(b) 4 α,7 β
(c) 4 α,4 β
(d) 4 α,1 β
Correct Answer:
4 α,1 β
Explanation:
α-particle = ₂He⁴, β-particle = ₋₁β and Nucleus = zXᴬ
Change in A occurs only due to α-emission.
Change in A = 228 – 212 = 16
This change is due to 4 α.
Again change in Z = 90 – 83 = 7
Change in Z due to 4α = 8
.·. Change in Z due to β = 8 – 7 = 1
This is due to one β.
Hence particles emitted = 4α, 1β.
Related Questions: - A radiactive substance emits 100 beta particles in the first 2s and 50 beta
- The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma
- If two electric bulbs have 40 W and 60 W ratings at 220 V, then the ratio
- If the nucleus ²⁷₁₃Al has nuclear radius of about 3.6 fm
- A beam of light contains two wavelengths 6500Å and 5200Å.
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A radiactive substance emits 100 beta particles in the first 2s and 50 beta
- The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma
- If two electric bulbs have 40 W and 60 W ratings at 220 V, then the ratio
- If the nucleus ²⁷₁₃Al has nuclear radius of about 3.6 fm
- A beam of light contains two wavelengths 6500Å and 5200Å.
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply