Question

write a balanced nuclear equation for the decay of each of the following nuclides to produce a beta particle
a. $^{136}_{53}\text{i}$
b. $^{133}_{51}\text{sb}$
c. $^{117}_{49}\text{in}$
write a balanced nuclear equation for the decay of each of the following nuclides to produce an alpha particle.
a. $^{226}_{88}\text{ra}$
b. $^{222}_{86}\text{rn}$
c. $^{239}_{94}\text{pu}$
d. $^{8}_{4}\text{be}$
electron configurations

Explanation:

Step1: Recall Beta Decay Rules

Beta decay (\( \beta^- \)): A neutron turns to proton, emits \( \beta^- \) (\( _{-1}^0e \)) and antineutrino. Mass number (\( A \)) stays same, atomic number (\( Z \)) increases by 1.

Part a: \( _{53}^{136}\text{I} \) Beta Decay

• Original: \( A = 136 \), \( Z = 53 \)
• After decay: \( Z = 53 + 1 = 54 \) (element Xe), \( A = 136 \)
• Equation: \( _{53}^{136}\text{I}

ightarrow _{54}^{136}\text{Xe} + _{-1}^0e \)

Part b: \( _{51}^{133}\text{Sb} \) Beta Decay

• \( Z = 51

ightarrow 52 \) (element Te), \( A = 133 \)

• Equation: \( _{51}^{133}\text{Sb}

ightarrow _{52}^{133}\text{Te} + _{-1}^0e \)

Part c: \( _{49}^{117}\text{In} \) Beta Decay

• \( Z = 49

ightarrow 50 \) (element Sn), \( A = 117 \)

• Equation: \( _{49}^{117}\text{In}

ightarrow _{50}^{117}\text{Sn} + _{-1}^0e \)

Step2: Recall Alpha Decay Rules

Alpha decay: Emits \( \alpha \)-particle (\( _2^4\text{He} \)). Mass number decreases by 4, atomic number decreases by 2.

Part a: \( _{88}^{226}\text{Ra} \) Alpha Decay

• \( A = 226 - 4 = 222 \), \( Z = 88 - 2 = 86 \) (element Rn)
• Equation: \( _{88}^{226}\text{Ra}

ightarrow _{86}^{222}\text{Rn} + _2^4\text{He} \)

Part b: \( _{86}^{222}\text{Rn} \) Alpha Decay

• \( A = 222 - 4 = 218 \), \( Z = 86 - 2 = 84 \) (element Po)
• Equation: \( _{86}^{222}\text{Rn}

ightarrow _{84}^{218}\text{Po} + _2^4\text{He} \)

Part c: \( _{94}^{239}\text{Pu} \) Alpha Decay

• \( A = 239 - 4 = 235 \), \( Z = 94 - 2 = 92 \) (element U)
• Equation: \( _{94}^{239}\text{Pu}

ightarrow _{92}^{235}\text{U} + _2^4\text{He} \)

Part d: \( _4^8\text{Be} \) Alpha Decay

• \( A = 8 - 4 = 4 \), \( Z = 4 - 2 = 2 \) (element He)
• Equation: \( _4^8\text{Be}

ightarrow _2^4\text{He} + _2^4\text{He} \) (since \( _2^4\text{He} \) is alpha, two alphas here)

Answer:

s (Balanced Equations):

Beta Decay:

a. \( \boldsymbol{_{53}^{136}\text{I}
ightarrow _{54}^{136}\text{Xe} + _{-1}^0e} \)
b. \( \boldsymbol{_{51}^{133}\text{Sb}
ightarrow _{52}^{133}\text{Te} + _{-1}^0e} \)
c. \( \boldsymbol{_{49}^{117}\text{In}
ightarrow _{50}^{117}\text{Sn} + _{-1}^0e} \)

Alpha Decay:

a. \( \boldsymbol{_{88}^{226}\text{Ra}
ightarrow _{86}^{222}\text{Rn} + _2^4\text{He}} \)
b. \( \boldsymbol{_{86}^{222}\text{Rn}
ightarrow _{84}^{218}\text{Po} + _2^4\text{He}} \)
c. \( \boldsymbol{_{94}^{239}\text{Pu}
ightarrow _{92}^{235}\text{U} + _2^4\text{He}} \)
d. \( \boldsymbol{_4^8\text{Be}
ightarrow _2^4\text{He} + _2^4\text{He}} \)