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KVPY SA Exam 2011 Question Paper
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© examsnet.com
Question : 29
Total: 80
A nucleus of lead
P
b
82
214
emits two electrons followed by an alpha particle. The resulting nucleus w ill have
82 protons and 128 neutrons
80 protons and 130 neutrons
82 protons and 130 neutrons
78 protons and 134 neutrons
Validate
Solution:
Pb
214
82
–
–
–
–
–
–
–
–
▶
2
e
−
+
2
He
4
X
210
82
82 → Proton
210 – 82 = 128 Neutron
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