Question

The missing particle in the reaction :
${}_{99}^{253}Es{+}_2^{4}He{\to }_{101}^{256}Md+\underset{–}{}$

• A. deuteron
• B. proton
• C. $\beta$ - particle
• D. neutron

Answer 1

Answer: (D)

Let the particle be $X$.
On summing the mass number on left hand side we get $253+4=257$ units.
On right hand side we have $256$ units only. So, we should have $1$ unit mass number on $X$ element.
On summing the atomic number on left hand side we get $99+2=101$ units
On right hand side we have $101$ units of atomic number. So, we should have $0$ unit atomic number on $X$ element.
So ${}_0^{1}X{=}_0^{1}n$ i.e. neutron.