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Binding Energy

When Uranium-...

Question

When Uranium- $\cdot$ $238$ undergoes alpha decay and then one beta decay the resulting isotope is :

T h - 234

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U - 234

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P a - 234

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T h - 230

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R a - 226

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Solution

The correct option is E $P a - 234$
When $U r a n i u m - 238$ undergoes alpha decay and then one beta decay the resulting isotope is $P a - 234$
$_{92}^{238} U \to_{90}^{234} T h +_{2}^{4} H e \to_{91}^{234} P a +_{- 1}^{0} β$

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Similar questions

_{92}^{238} U

_{90}^{234} T h,

_{91}^{234} P a,

_{92}^{234} U,

_{90}^{230}

T h

_{88}^{226} R a

_{92} U^{238} \to_{90} T h^{234} \to_{91} P a^{234} \to

_{92} U^{234} \to_{90} T h^{230} \to_{88} R a^{226}

_{92} U^{238}

_{90} {Th}^{234}

_{91} {Pa}^{234}

_{92} U^{234}

_{90} {Th}^{230}

_{88} {Ra}^{226}

_{92}^{238} U - x_{1} - - \to_{90}^{234} T h - x_{2} - - \to_{91}^{234} P a - x_{3} - - \to^{234} Z - x_{4} - - \to_{90}^{230} T h x_{1}, x_{2}, x_{3} a n d x_{4}

_{92}^{238} U

_{90}^{234} T h

_{2}^{4} H e

_{92}^{238} U

_{90}^{234} T h

_{92}^{238} U \to_{90}^{234} T h +_{2}^{4} H e + e n e r g y

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