Question

Which among the following are the correct statements?

• A. $C{O}_2$ and $N_{2}O$ are called isostere
• B. The first artificial transmutation was made by Rutherford
• C. A certain radioactive isotope decay from ${}_{92}{U}^{235}$ ends up in ${}_{82}P{b}^{207}$ by the emission of $7\alpha$ and $4\beta$ particles
• D. During disintegration, ${}_{89}A{c}^{228}$ appears in $4n$ series

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $C{O}_2$ and $N_{2}O$ are called isostere
B The first artificial transmutation was made by Rutherford
C During disintegration, ${}_{89}A{c}^{228}$ appears in $4n$ series
D A certain radioactive isotope decay from ${}_{92}{U}^{235}$ ends up in ${}_{82}P{b}^{207}$ by the emission of $7\alpha$ and $4\beta$ particles

In $C{O}_2$ and $N_{2}O$, there are 3 atoms and 22 electrons, hence, they are isostere.
The first artificial transmutation was made by Rutherford is a correct statement.

Losing an $\alpha$-particle shows a loss of 4 units of its mass and 2 units in its atomic number. Losing a $\beta$-particle shows a gain in its atomic number by 1 unit, but the mass number remains the same.

${}_{92}{U}^{235}-7{(}_{2}H{e}^4)-4{(}_{-1}{e}^0){\to }_{82}{U}^{207}$

${}_{89}A{c}^{228}=\frac{Massnumber}{4}=57.0$, divisible by 4, hence, a member of $4n$ series.