Question

1. write a nuclear equation for the alpha decay of ²¹¹pa.
2. write a nuclear equation for the beta decay of ²²³fr.
3. write a nuclear equation for the alpha decay of ¹⁴⁹sm.
4. write a nuclear equation for the beta decay of ¹⁶⁵pm.
5. write a nuclear equation for the alpha decay of ²⁴⁹md.
6. write a nuclear equation for the alpha decay of ¹⁴⁶sm.
7. write a nuclear equation for the beta decay of ¹⁹⁸at.
8. write a nuclear equation for the alpha decay of ¹⁵⁰gd.
9. write a nuclear equation for the beta decay of ¹⁵²xe.
10. write a nuclear equation for the beta decay of ¹²⁰cs.

Explanation:

Step1: Recall alpha - decay formula

In alpha - decay, the parent nucleus emits an alpha particle ($_{2}^{4}$He). The mass number of the parent nucleus decreases by 4 and the atomic number decreases by 2.

Step2: Recall beta - decay formula

In beta - decay ($\beta^{-}$), a neutron in the nucleus is converted into a proton and an electron. The electron is emitted as a beta particle ($_{ - 1}^{0}$e). The mass number of the parent nucleus remains the same and the atomic number increases by 1.

1. For the alpha decay of $_{91}^{211}$Pa:

$_{91}^{211}$Pa $
ightarrow$ $_{89}^{207}$Ac + $_{2}^{4}$He

1. For the beta decay of $_{87}^{223}$Fr:

$_{87}^{223}$Fr $
ightarrow$ $_{88}^{223}$Ra + $_{ - 1}^{0}$e

1. For the alpha decay of $_{62}^{149}$Sm:

$_{62}^{149}$Sm $
ightarrow$ $_{60}^{145}$Nd + $_{2}^{4}$He

1. For the beta decay of $_{61}^{165}$Pm:

$_{61}^{165}$Pm $
ightarrow$ $_{62}^{165}$Sm + $_{ - 1}^{0}$e

1. For the alpha decay of $_{101}^{248}$Md:

$_{101}^{248}$Md $
ightarrow$ $_{99}^{244}$Es + $_{2}^{4}$He

1. For the alpha decay of $_{62}^{148}$Sm:

$_{62}^{148}$Sm $
ightarrow$ $_{60}^{144}$Nd + $_{2}^{4}$He

1. For the beta decay of $_{85}^{198}$At:

$_{85}^{198}$At $
ightarrow$ $_{86}^{198}$Rn + $_{ - 1}^{0}$e

1. For the alpha decay of $_{64}^{150}$Gd:

$_{64}^{150}$Gd $
ightarrow$ $_{62}^{146}$Sm + $_{2}^{4}$He

1. For the beta decay of $_{54}^{152}$Xe:

$_{54}^{152}$Xe $
ightarrow$ $_{55}^{152}$Cs + $_{ - 1}^{0}$e

1. For the beta decay of $_{55}^{120}$Cs:

$_{55}^{120}$Cs $
ightarrow$ $_{56}^{120}$Ba + $_{ - 1}^{0}$e

Answer:

1. $_{91}^{211}$Pa $

ightarrow$ $_{89}^{207}$Ac + $_{2}^{4}$He

1. $_{87}^{223}$Fr $

ightarrow$ $_{88}^{223}$Ra + $_{ - 1}^{0}$e

1. $_{62}^{149}$Sm $

ightarrow$ $_{60}^{145}$Nd + $_{2}^{4}$He

1. $_{61}^{165}$Pm $

ightarrow$ $_{62}^{165}$Sm + $_{ - 1}^{0}$e

1. $_{101}^{248}$Md $

ightarrow$ $_{99}^{244}$Es + $_{2}^{4}$He

1. $_{62}^{148}$Sm $

ightarrow$ $_{60}^{144}$Nd + $_{2}^{4}$He

1. $_{85}^{198}$At $

ightarrow$ $_{86}^{198}$Rn + $_{ - 1}^{0}$e

1. $_{64}^{150}$Gd $

ightarrow$ $_{62}^{146}$Sm + $_{2}^{4}$He

1. $_{54}^{152}$Xe $

ightarrow$ $_{55}^{152}$Cs + $_{ - 1}^{0}$e

1. $_{55}^{120}$Cs $

ightarrow$ $_{56}^{120}$Ba + $_{ - 1}^{0}$e