The following 93 237 N p { }_{93}^{237} \mathrm{~Np} 93 237 Np ? 91 233 P a { }_{91}^{233} \mathrm{~Pa} 91 233 Pa can be balanced how?A) 4 2 H { }_{4}^{2} \mathrm{H} 4 2 H B) 2 4 H e { }_{2}^{4} \mathrm{He} 2 4 He C) 2 4 H { }_{2}^{4} \mathrm{H} 2 4 H .D) 2 4 H { }_{2}^{4} \mathrm{H} 2 4 H

Question

The following   93 237   N p { }_{93}^{237} \mathrm{~Np} 93 237 ​   Np   ?   91 233   P a { }_{91}^{233} \mathrm{~Pa} 91 233 ​   Pa   can be balanced how?A)    4 2 H { }_{4}^{2} \mathrm{H} 4 2 ​ H B)    2 4 H e { }_{2}^{4} \mathrm{He} 2 4 ​ He C)    2 4 H { }_{2}^{4} \mathrm{H} 2 4 ​ H  .D)    2 4 H { }_{2}^{4} \mathrm{H} 2 4 ​ H

Zybrea Knight · Accepted Answer

Final Answer : B, Explanation :   The correct answer is B)   2 4 H e { }_{2}^{4} \mathrm{He} 2 4 ​ He  , which represents an alpha particle. In nuclear reactions, the sum of mass numbers and the sum of atomic numbers must be equal on both sides of the equation. Here,   93 237 N p { }_{93}^{237} \mathrm{Np} 93 237 ​ Np   decays into   91 233 P a { }_{91}^{233} \mathrm{Pa} 91 233 ​ Pa   and an alpha particle   2 4 H e { }_{2}^{4} \mathrm{He} 2 4 ​ He  , balancing both the atomic and mass numbers.

The following $}_{93}^{237} \mathrm{~Np}$ ? $}_{91}^{233} \mathrm{~Pa}$ can be balanced how?

A) $}_{4}^{2} \mathrm{H}$

B) $}_{2}^{4} \mathrm{He}$

C) $}_{2}^{4} \mathrm{H}$ .

D) $}_{2}^{4} \mathrm{H}$

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Learning Objectives

Learning Objectives

Related questions

The following 93237 Np{ }_{93}^{237} \mathrm{~Np}93237​ Np ? 91233 Pa{ }_{91}^{233} \mathrm{~Pa}91233​ Pa can be balanced how?A) 42H{ }_{4}^{2} \mathrm{H}42​H B) 24He{ }_{2}^{4} \mathrm{He}24​He C) 24H{ }_{2}^{4} \mathrm{H}24​H .D) 24H{ }_{2}^{4} \mathrm{H}24​H

Definitions

Learning Objectives

Learning Objectives

Related questions

The following $}_{93}^{237} \mathrm{~Np}$ ? $}_{91}^{233} \mathrm{~Pa}$ can be balanced how?

A) $}_{4}^{2} \mathrm{H}$

B) $}_{2}^{4} \mathrm{He}$

C) $}_{2}^{4} \mathrm{H}$ .

D) $}_{2}^{4} \mathrm{H}$