A radioactive element _ 90 X^ 238 decays into _ 83 Y^ 222 . The number of β -particles emitted are

Correct answer: A. 1 α -particles are emitted =(238-222)/(4)=4 The atomic number is decreased 90-4 × 2=82 As atomic number of _ 83 Y^ 222 , So, atomic number is increased by 1 , therefore, one β -particle is emitted.

JIPMER 2013

Exam: JIPMER
Year: 2013
Subject: Physics
Chapter: Nuclear Physics

JIPMER 2013 · Physics · Nuclear Physics

A radioactive element _ 90 X^ 238 decays into _ 83 Y^ 222 . The...

A radioactive element

{ }_{90} X^{238}

decays into

{ }_{83} Y^{222}

. The number of

\beta

-particles emitted are

Options

Correct answer

A. 1

Step-by-step solution

\alpha

-particles are emitted

=\frac{238-222}{4}=4

The atomic number is decreased

90-4 \times 2=82

As atomic number of

{ }_{83} Y^{222}

, So, atomic number is increased by 1 , therefore, one

\beta

-particle is emitted.

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