# Which missing item would complete this alpha decay reaction?

## $\text{_(98)^(251) "Cf" -> ""_(96)^(247) "Cm" + "___}$ $A .$ $\text{_(-1)^(0) "e}$ $B .$ $\text{_(97)^(247) "Bk}$ $C .$ $\text{_(2)^(4) "He}$ $D .$ $\text{_(92)^(238) "U}$ $E .$ ""_(0)^(0) gamma

Feb 3, 2017

${\text{Option" C-> }}_{2}^{4} H e$

#### Explanation:

$\text{The decay would have helium left after the reaction.}$

${\text{_98^251 Cf → ""_96^247Cm +}}_{2}^{4} H e$

$\text{We know the atomic masses of the elements but we do not know}$
$\text{know the number of protons of the substances .}$
$\text{ But for this reaction number of neutrons are also required }$

First determine the number of neutrons in Cf

$\text{no. of neutrons = atomic weight - no. of protons}$

$\text{251 - 98 = 153 neutrons}$

Then determine the number of neutrons in Cf

$\text{no. of neutrons = atomic weight - no. of protons}$

$\text{247 - 96 = 151 neutrons}$

This means there are more 2 protons and neutrons

$\text{Decay in neutrons}$

98 → 96 + 2

$\text{Decay in protons}$

153 → 151 + 2  helium atom as the electrons are number of electrons = no. of protons

Energy can be neither created nor destroyed .

Protons and neutrons are energy in scientific words

As energy cannot be destroyed so 251 → 247 cannot happen and there should be something more .

This is an alpha decay equation so this means there are more 2 electrons and neutrons which mean that the element is Helium without its electrons.An
α
particle is 2 protons and 2 neutrons bound together like they are within the nucleus of a Helium atom; it is so like a Helium nucleus that it is often written as $\frac{4}{2} H e 2 +$