Question

In transition 'a', three \(\alpha\) particles and two \(\beta\) particles are emitted in succession from the element \({}_{86}X^{226}\). In transition 'b', two gamma rays, one \(\beta\) particle and two \(\alpha\) particles are emitted in succession from element \({}_{96}Y^{236}\). The atomic number and the mass number of the new elements formed respectively are

• a) for first transition:- atomic number \(=83\), mass number \(=216\) b) for the second transition:- atomic number \(=82\), mass number \(=214\)
• a) for first transition:- atomic number \(=80\), mass number \(=225\) b) for the second transition:- atomic number \(=90\), mass number \(=220\)
• a) for first transition:- atomic number \(=82\), mass number \(=214\) b) for the second transition:- atomic number \(=93\), mass number \(=228\)
• a) for first transition:- atomic number \(=80\), mass number \(=214\) b) for the second transition:- atomic number \(=90\), mass number \(=226\)

Hint:

An alpha particle lowers Z by 2 and A by 4, a beta particle raises Z by 1, and gamma rays change nothing.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Alpha emission: \(Z\) decreases by 2 and \(A\) decreases by 4.
Beta emission: \(Z\) increases by 1 and \(A\) is unchanged.
Gamma emission changes neither \(Z\) nor \(A\).

Step 2: Transition a
Start with \({}_{86}X^{226}\), three alpha and two beta.
\(A=226-3\times4=214\)
\(Z=86-3\times2+2=82\)

Step 3: Transition b
Start with \({}_{96}Y^{236}\), two alpha, one beta and two gamma.
\(A=236-2\times4=228\)
\(Z=96-2\times2+1=93\)

Step 4: Compare the options
The pair (82, 214) and (93, 228) matches option (C).

Final Answer:
The new elements have Z=82, A=214 and Z=93, A=228. \[ \boxed{\text{(C) }\text{a) Z = 82, A = 214; b) Z = 93, A = 228}} \]