A radioactive element 90X238 decays into 83 Upsilon 222. The number of β-particle emitted are:

Question

A radioactive element 90X238 decays into 83 Upsilon 222. The number of β-particle emitted are: (A) 1 (B) 2 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

α -particle are emitted =(238-222/4)=4 The atomic number is decreased 90-4× 2=82 As atomic number of 83Y222, so atomic number is increased by 1, therefore one β -particle is emitted.

Q. A radioactive element $_{90} X^{238}$ decays into $_{83} Υ^{222} .$ The number of $β$ -particle emitted are:

1

2

4

6

$α$ -particle 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 $β$ -particle is emitted.

Q. A radioactive element 90​X238 decays into 83​Υ222. The number of β-particle emitted are:

1

2

4

6

α -particle are emitted =4238−222​=4 The atomic number is decreased 90−4×2=82 As atomic number of 83​Y222, so atomic number is increased by 1, therefore one β -particle is emitted.

Q. A radioactive element $_{90} X^{238}$ decays into $_{83} Υ^{222} .$ The number of $β$ -particle emitted are:

$α$ -particle 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 $β$ -particle is emitted.