$T h^{234}$ disintegrates and emits 6β- and 7α-particles to form a stable element. Find the atomic number and mass number of the stable product. Also identify the element.
(IIT-JEE, 2004)

_{82} P b^{206} .

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_{80} P b^{204} .

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_{82} P b^{204} .

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_{80} H g^{202} .

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Solution

The correct option is A $_{82} P b^{206} .$
Nuclear reaction is:

$_{90} T h^{234} \to 7_{2} H e^{4} + 6_{- 1} e^{0} +_{z} X^{A}$

Equating atomic number on both sides

90 = 14 + 6 x (-1) + Z

∴ Z = 82

Equating mass number on both sides

238 = 7 x 4 + 6 x 0 + A

∴ A = 206
Thus, element with atomic number 82 and mass number 206 is $_{82} P b^{206} .$

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Th234 disintegrates and emits 6β- and 7α-particles to form a stable element. Find the atomic number and mass number of the stable product. Also identify the element. (IIT-JEE, 2004)

The correct option is A 82Pb206.Nuclear reaction is: 90Th234→7 2He4+6−1e0+zXA Equating atomic number on both sides 90 = 14 + 6 x (-1) + Z ∴ Z = 82 Equating mass number on both sides 238 = 7 x 4 + 6 x 0 + A ∴ A = 206 Thus, element with atomic number 82 and mass number 206 is 82Pb206.

$T h^{234}$ disintegrates and emits 6β- and 7α-particles to form a stable element. Find the atomic number and mass number of the stable product. Also identify the element.
(IIT-JEE, 2004)