Step 1: Effect of Alpha ( \( \alpha \) ) Decay
An alpha particle (\( \alpha \)) has a mass number of 4 and an atomic number of 2.
When element \( X \) undergoes alpha decay:
$$ {^{290}_{82}X -> ^{286}_{80}Y + ^{4}_{2}\alpha} $$
Thus, the new nucleus \( Y \) has:
Mass number = \( 290 - 4 = 286 \)
Atomic number = \( 82 - 2 = 80 \)
Step 2: Effect of Positron (\( e^+ \)) Emission
Positron emission (\( \beta^+ \)) decreases the atomic number by 1 without changing the mass number:
$$ {^{286}_{80}Y -> ^{286}_{79}Z + e^+} $$
Thus, \( Z \) has:
Mass number = 286 (unchanged)
Atomic number = \( 80 - 1 = 79 \)
Step 3: Effect of Beta-Minus (\( \beta^- \)) Decay
Beta-minus (\( \beta^- \)) emission increases the atomic number by 1 without affecting the mass number:
$$ {^{286}_{79}Z -> ^{286}_{80}P + \beta^-} $$
Thus, \( P \) has:
Mass number = 286 (unchanged)
Atomic number = \( 79 + 1 = 80 \)
Step 4: Effect of Electron Capture (\( e^- \))
Electron capture decreases the atomic number by 1, keeping the mass number unchanged:
$$ {^{286}_{80}P + e^- -> ^{286}_{81}Q} $$
Thus, \( Q \) has:
Mass number = 286 (unchanged)
Atomic number = \( 80 + 1 = 81 \)
Conclusion
The final product \( Q \) has mass number 286 and atomic number 81, which corresponds to option (4).
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The potential (V) at any axial point, at 2 m distance(r) from the centre of the dipole of dipole moment vector
\(\vec{P}\) of magnitude, 4 × 10-6 C m, is ± 9 × 103 V.
(Take \(\frac{1}{4\pi\epsilon_0}=9\times10^9\) SI units)
Reason R : \(V=±\frac{2P}{4\pi \epsilon_0r^2}\), where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below :
The output (Y) of the given logic gate is similar to the output of an/a :
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |