The atomic number A and mass number M of the nuclide formed when three alpha - and two - particles are emitted from 23892238U

The correct option is D A = 88, M = 226 _92^238U → 3 _2^4He + 2 β^ + _A^MX Mass balance: #160; #160; #160; #160; #160; #160; #160; ...

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Standard IX

Chemistry

Atomic Number

The atomic nu...

Question

The atomic number ( $A$ ) and mass number ( $M$ ) of the nuclide formed when three alpha $α$ and two $β$ particles are emitted from $238_{92}^{238} U$

A = 87, M = 233

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A = 86, M = 226

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A = 88, M = 227

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A = 88, M = 226

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Solution

The correct option is D $A = 88, M = 226$
$_{92}^{238} U \to 3_{2}^{4} H e + 2 β^{-} +_{A}^{M} X$
Mass balance:
$238 = 12 + 0 + M$
$M = 226$
Charge balance (i.e., proton balance)
$92 = 6 - 2 + A$
$A = 88$

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Similar questions

Find the Q-value and the kinetic energy of the emitted

α

- Particle in the

α

-decay of (a)

_{86}^{220} R n

and (b)

_{88}^{226} R a

Given:

m (_{88}^{226} R a) = 226.02540 u

m (_{86}^{222} R n) = 222.01750 u

m (_{84}^{216} P o) = 216.00189 u

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) and (b).

Given = 226.02540 u, = 222.01750 u,

= 220.01137 u, = 216.00189 u.

Q. Find whether

α

- decay or any of the

β -

decay are allowed for

_{89}^{226} A c

. Given masses are:

M (_{89}^{226} A c) = 226.028356 a m u

M (_{87}^{222} F r) = 222.017415 a m u

M (_{90}^{226} T h) = 226.017388 a m u

M (_{88}^{226} R a) = 226.025406 a m u

M (_{2}^{4} H e) = 4.002603 a m u

A nuclide A (with mass number m and atomic number n) disintegrates emitting an alpha and one beta particle. The resulting nuclide B has mass number and atomic number respectively equal to

Q. When

_{88}^{226} R a

emits an

α

-particle, the new element formed belongs to:

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The atomic number (A) and mass number (M) of the nuclide formed when three alpha α and two β particles are emitted from 23823892U

The correct option is D A=88,M=22623892U→342He+2β−+MAXMass balance: 238=12+0+M M=226Charge balance (i.e., proton balance) 92=6−2+A A=88

The atomic number ( $A$ ) and mass number ( $M$ ) of the nuclide formed when three alpha $α$ and two $β$ particles are emitted from $238_{92}^{238} U$

The correct option is D $A = 88, M = 226$
$_{92}^{238} U \to 3_{2}^{4} H e + 2 β^{-} +_{A}^{M} X$
Mass balance:
$238 = 12 + 0 + M$
$M = 226$
Charge balance (i.e., proton balance)
$92 = 6 - 2 + A$
$A = 88$