If _ 92 U^ 238 emits 8 α -particles and 6 β -particles, then the resulting nucleus is

Correct answer: B. _ 82 ~Pb^ 206 After one α -emission, the daughter nucleus reduces in mass number by 4 unit and in atomic number by 2 unit. In β -emission the atomic number of daughter nucleus increases by 1 unit. The reaction can be written as _ 92 U^ 238 -8 α _ 76 X^ 206 -6 β _ 82 Y^ 206 Thus, the resulting nucleus is _ 82 Y^ 206 , i e, _ 82 ~Pb^ 206 .

JIPMER 2010

Exam: JIPMER
Year: 2010
Subject: Physics
Chapter: Nuclear Physics

JIPMER 2010 · Physics · Nuclear Physics

If _ 92 U^ 238 emits 8 α -particles and 6 β -particles, then the...

{ }_{92} \mathrm{U}^{238}

emits

8 \alpha

-particles and

6 \beta

-particles, then the resulting nucleus is

Options

A ${ }_{82} \mathrm{U}^{206}$
B ${ }_{82} \mathrm{~Pb}^{206}$
C ${ }_{82} \mathrm{U}^{210}$
D ${ }_{82} \mathrm{U}^{214}$

Correct answer

B. ${ }_{82} \mathrm{~Pb}^{206}$

Step-by-step solution

After one

\alpha

-emission, the daughter nucleus reduces in mass number by 4 unit and in atomic number by 2 unit. In

\beta

-emission the atomic number of daughter nucleus increases by 1 unit. The reaction can be written as

{ }_{92} \mathrm{U}^{238} \xrightarrow{-8 \alpha}{ }_{76} X^{206} \xrightarrow{-6 \beta}{ }_{82} Y^{206}

Thus, the resulting nucleus is

{ }_{82} Y^{206}, i e,{ }_{82} \mathrm{~Pb}^{206}

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