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At \(n=1\), \(E\) is most negative (\(-2.178 \times 10^{-18}\) J for H). Most negative = most tightly bound (lowest energy state). Saying it is loosely bound is WRONG โ it is the most tightly bound state.
The other statements are all correct interpretations of the Bohr energy equation. The electron at \(n=1\) is hardest to remove (highest ionisation energy).
How many values of $k$ allow the system of equations $(k+1)x + 8y = 4k$ and $kx + (k+3)y = 3k-1$ to have no solution?
For no solution, the determinant of the coefficient matrix must be zero but the system must be inconsistent.
$\Delta = (k+1)(k+3) - 8k = k^2+4k+3-8k = k^2-4k+3 = (k-1)(k-3) = 0$
So $k=1$ or $k=3$. Check each:
$k=1$: Equations become $2x+8y=4$ and $x+4y=2$ โ same line (infinite solutions). Not "no solution".
$k=3$: Equations become $4x+8y=12$ and $3x+6y=8$. Simplify: $x+2y=3$ and $x+2y=8/3$ โ contradiction! No solution.
So exactly $\mathbf{1}$ value of $k$ gives no solution.
What is the minimum energy needed to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ into a circular orbit at altitude $2R$?
The orbit radius is $r = R + 2R = 3R$.
Orbital KE: $KE = \frac{GmM}{2r} = \frac{GmM}{6R}$
Total mechanical energy in orbit: $E_{orbit} = -\frac{GmM}{2r} = -\frac{GmM}{6R}$
Energy on surface: $E_{surface} = -\frac{GmM}{R}$
Minimum energy required:
$\Delta E = E_{orbit} - E_{surface} = -\frac{GmM}{6R} + \frac{GmM}{R} = \frac{GmM}{R}\left(1 - \frac{1}{6}\right) = \frac{5GmM}{6R}$
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