?  + _2^4He + Q (ii) "_11^22Na ->?  + _10^22Ne + v - Physics | Shaalaa.com" />?  + _2^4He + Q (ii) "_11^22Na ->?  + _10^22Ne + v " />?  + _2^4He + Q (ii) "_11^22Na ->?  + _10^22Ne + v, Nuclear Energy - Nuclear Reactor" />
हिंदी

Complete the following nuclear reactions for α and β deca : (i) "_92^238U ->?  + _2^4He + Q (ii) "_11^22Na ->?  + _10^22Ne + v - Physics

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प्रश्न

Complete the following nuclear reactions for α and β deca :

(i) `"_92^238U ->?  + _2^4He + Q`

(ii) `"_11^22Na ->?  + _10^22Ne + v`

टिप्पणी लिखिए

उत्तर

(i) By balancing the atomic number and mass number we can easily find that the product will be an element with atomic number 90 and mass number 234, i.e. Thorium.  
`"_92^238U ->  + _90^234Th + _2^4He + Q`

(ii) As we can see there is no change in mass number during the reaction but the value of Z reduces by factor 1, hence we can conclude there is a positron beta decay happening in the reaction, so the missing product will be a positive beta particle.
The balanced reaction is as follows:
`"_11^22Na -> _10^22Ne + _1^0beta + v`

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2018-2019 (March) 55/3/1

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