When - - 90 - - T - - 228 - transforms to - - 83 - - Bi - - 212 - then the number to the emitted alpha and beta particles is- respectively-8alpha -7beta 4alpha -7beta 4alpha -4beta 4alpha -1beta

Question

When - - 90 - - T - - 228 - transforms to  - - 83 - - Bi - - 212 - then the number to the emitted alpha  and beta  particles is- respectively-8alpha -7beta 4alpha -7beta 4alpha -4beta 4alpha -1beta

Toppr Admin · Accepted Answer

Z-90TA-228-x27F6-Z-x2032-83BiA-x2032-212Number of -x3B1- particles emittedn-x3B1-A-x2212-A-x2032-4-226-x2212-2124-4Number of -x3B2- particles emittedn-x3B2-2n-x3B1-x2212-Z-Z-x2032-2-xD7-4-x2212-90-83-1

When ${_{90} T}^{228}$ transforms to ${_{83} B i}^{212}$ , then the number to the emitted $α$ and $β$ particles is, respectively:
$8 α, 7 β$
$4 α, 7 β$
$4 α, 4 β$
$4 α, 1 β$

${_{Z = 90} T}^{A = 228} ⟶ {_{Z^{'} = 83} B i}^{A^{'} = 212}$
Number of $α$ particles emitted
$n_{α} = \frac{A - A^{'}}{4} = \frac{226 - 212}{4} = 4$
Number of $β$ particles emitted
$n_{β} = 2 n_{α} - Z + Z^{'} = 2 \times 4 - 90 + 83 = 1$

When 90T228 transforms to 83Bi212, then the number to the emitted α and β particles is, respectively:8α,7β4α,7β4α,4β4α,1β

Z=90TA=228⟶Z′=83BiA′=212Number of α particles emittednα=A−A′4=226−2124=4Number of β particles emittednβ=2nα−Z+Z′=2×4−90+83=1

When ${_{90} T}^{228}$ transforms to ${_{83} B i}^{212}$ , then the number to the emitted $α$ and $β$ particles is, respectively:
$8 α, 7 β$
$4 α, 7 β$
$4 α, 4 β$
$4 α, 1 β$

${_{Z = 90} T}^{A = 228} ⟶ {_{Z^{'} = 83} B i}^{A^{'} = 212}$
Number of $α$ particles emitted
$n_{α} = \frac{A - A^{'}}{4} = \frac{226 - 212}{4} = 4$
Number of $β$ particles emitted
$n_{β} = 2 n_{α} - Z + Z^{'} = 2 \times 4 - 90 + 83 = 1$