Complete the following reaction-92-238- U rightarrow -90-234- Th- - Energy-2-4- He4 -1-0- beta4 -0-1- NeutronNone of the above

Question

Complete the following reaction-92-238- U rightarrow -90-234- Th- - Energy-2-4- He4  -1-0- beta4  -0-1-  NeutronNone of the above

Toppr Admin · Accepted Answer

The Mass number of particle emitted is Mass-xA0-number-xA0-of-xA0-U-x2212-mass-xA0-number-xA0-of-xA0-Th-238-x2212-234-4-The Atomic number of particle emitted is Atomic-xA0-number-xA0-of-xA0-U-x2212-Atomic-xA0-number-xA0-of-xA0-Th-92-x2212-90-2-So the emitted particle is -x3B1- particle i-e- 42He

Complete the following reaction.

$_{92}^{238} U \to_{90}^{234} T h +$ ................ + Energy
$_{2}^{4} H e$
$4_{- 1}^{0} β$
$4_{0}^{1} N e u t r o n$
None of the above

The Mass number of particle emitted is $M a s s n u m b e r o f U - m a s s n u m b e r o f T h = 238 - 234 = 4$ .
The Atomic number of particle emitted is $A t o m i c n u m b e r o f U - A t o m i c n u m b e r o f T h = 92 - 90 = 2$ .
So the emitted particle is $α$ particle i.e., $_{2}^{4} H e$

Complete the following reaction.23892U→23490Th+................ + Energy42He4 0−1β4 10 NeutronNone of the above

The Mass number of particle emitted is Mass number of U−mass number of Th=238−234=4.The Atomic number of particle emitted is Atomic number of U−Atomic number of Th=92−90=2.So the emitted particle is α particle i.e., 42He

Complete the following reaction.

$_{92}^{238} U \to_{90}^{234} T h +$ ................ + Energy
$_{2}^{4} H e$
$4_{- 1}^{0} β$
$4_{0}^{1} N e u t r o n$
None of the above

The Mass number of particle emitted is $M a s s n u m b e r o f U - m a s s n u m b e r o f T h = 238 - 234 = 4$ .
The Atomic number of particle emitted is $A t o m i c n u m b e r o f U - A t o m i c n u m b e r o f T h = 92 - 90 = 2$ .
So the emitted particle is $α$ particle i.e., $_{2}^{4} H e$