The balanced nuclear equations for each of the following- -emission from bismuth-189

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Standard XII

Physics

Beta Decay

The balanced ...

Question

The balanced nuclear equations for each of the following:
$α$ -emission from bismuth-189

_{93}^{189} B i ⟶_{4}^{4} H e +_{81}^{185} T l

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_{83}^{189} B i ⟶_{2}^{4} H e +_{81}^{185} T l

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_{73}^{189} B i ⟶_{3}^{4} H e +_{81}^{185} T l

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None of these

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Solution

The correct option is D $_{83}^{189} B i ⟶_{2}^{4} H e +_{81}^{185} T l$
The reaction involving $α$ -emission from bismuth- $189$ is given below:

$_{83}^{189} B i ⟶_{2}^{4} H e +_{81}^{185} T l$

As in this case both atomic and mass number is conserved (sum is equal)

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The radioactive decay, $_{83} B i^{211} \to_{81} T l^{207},$ takes place in 10 L closed container at $0^{\circ} C .$ Half - life of $_{83} B i^{211}$ is 2.15 min. Starting with 1 mole of $_{83} B i^{211}$ , the pressure developed in the container after 4.30 min. will be

Q. Which of the following option correctly represent the correct set of stability order for the given ions?
(I)

G a^{3 +} > I n^{3 +} > T l^{3 +}

(II)

G a^{3 +} < I n^{3 +} < T l^{3 +}

(III)

G e^{+ 4} > S n^{+ 4} > P b^{+ 4}

(IV)

G e^{+ 2} < S n^{2 +} < P b^{2 +}

(V)

A s^{3 +} < S b^{3 +} < B i^{3 +}

What is the standard potential of the $T l^{3 +} / T l$ electrode?

$T l^{3 +} + 2 e^{-} \to T l^{+} E^{0} = 1.26 V T l^{+} + e^{-} \to T l E^{0} = - 0.336 V$

Q. What is the standard potential of the

T l^{3 +} / T l

electrode?

T l^{3 +} + 2 e^{-} \to T l^{+}; E^{\circ} = 1.26 v o l t

T l^{+} + e^{-} \to T l; E^{\circ} = - 0.336 v o l t

$T l + H_{2} O + O_{2} \to A$ . A is

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The balanced nuclear equations for each of the following:α-emission from bismuth-189

The correct option is D 18983Bi⟶42He+18581TlThe reaction involving α-emission from bismuth-189 is given below:18983Bi⟶42He+18581TlAs in this case both atomic and mass number is conserved (sum is equal)

The balanced nuclear equations for each of the following:
$α$ -emission from bismuth-189

The correct option is D $_{83}^{189} B i ⟶_{2}^{4} H e +_{81}^{185} T l$
The reaction involving $α$ -emission from bismuth- $189$ is given below:

$_{83}^{189} B i ⟶_{2}^{4} H e +_{81}^{185} T l$

As in this case both atomic and mass number is conserved (sum is equal)