Question:

82290XαYe+ZβPeQ^{290}_{82}X\stackrel{\alpha}{\rightarrow}Y\stackrel{e^+}{\rightarrow}Z\stackrel{\beta^-}{\rightarrow}P\stackrel{e^-}{\rightarrow}Q
In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are :

Updated On: Apr 3, 2025
  • 280, 81
  • 286, 80
  • 288, 82
  • 286, 81
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The Correct Option is D

Solution and Explanation

Step 1: Effect of Alpha ( α \alpha ) Decay

An alpha particle (α \alpha ) has a mass number of 4 and an atomic number of 2.

When element X X undergoes alpha decay:

82290X>80286Y+24α {^{290}_{82}X -> ^{286}_{80}Y + ^{4}_{2}\alpha}

Thus, the new nucleus Y Y has:

Mass number = 2904=286 290 - 4 = 286

Atomic number = 822=80 82 - 2 = 80

Step 2: Effect of Positron (e+ e^+ ) Emission

Positron emission (β+ \beta^+ ) decreases the atomic number by 1 without changing the mass number:

80286Y>79286Z+e+ {^{286}_{80}Y -> ^{286}_{79}Z + e^+}

Thus, Z Z has:

Mass number = 286 (unchanged)

Atomic number = 801=79 80 - 1 = 79

Step 3: Effect of Beta-Minus (β \beta^- ) Decay

Beta-minus (β \beta^- ) emission increases the atomic number by 1 without affecting the mass number:

79286Z>80286P+β {^{286}_{79}Z -> ^{286}_{80}P + \beta^-}

Thus, P P has:

Mass number = 286 (unchanged)

Atomic number = 79+1=80 79 + 1 = 80

Step 4: Effect of Electron Capture (e e^- )

Electron capture decreases the atomic number by 1, keeping the mass number unchanged:

80286P+e>81286Q {^{286}_{80}P + e^- -> ^{286}_{81}Q}

Thus, Q Q has:

Mass number = 286 (unchanged)

Atomic number = 80+1=81 80 + 1 = 81

Conclusion

The final product Q Q has mass number 286 and atomic number 81, which corresponds to option (4).

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