Grade 11 : Chemistry - Atom Structure (Set 3)

Explanation:Using Planck's photon energy equation $E = \frac{hc}{\lambda}$, where $\lambda = 300\text{ nm} = 3.0 \times 10^{-7}m$:
$$E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{3.0 \times 10^{-7}} = 6.626 \times 10^{-19}\text{ J}$$
Rounding to three significant figures gives $6.63 \times 10^{-19}\text{ J}$.

Explanation:The radius of a Bohr orbit is given by $r_n = 0.529 \frac{n^2}{Z}\text{ \AA}$.
For $\text{He}^+$ ($n=2, Z=2$): $r_2(\text{He}^+) = 0.529 \times \frac{2^2}{2} = 2 \times 0.529\text{ \AA}$.
For $\text{H}$ atom ($n=1, Z=1$): $r_1(\text{H}) = 0.529 \times \frac{1^2}{1} = 1 \times 0.529\text{ \AA}$.
Ratio = $\frac{2 \times 0.529}{1 \times 0.529} = 2 : 1$.

Explanation:Energy levels for Hydrogen are given by $E_n = -\frac{13.6}{n^2}\text{ eV}$.
Ground state ($n=1$): $E_1 = -13.6\text{ eV}$.
First excited state ($n=2$): $E_2 = -\frac{13.6}{4} = -3.4\text{ eV}$.
Excitation energy $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = +10.2\text{ eV}$.

Explanation:The total number of spectral lines emitted during transitions to the ground state is given by:
$$N = \frac{\Delta n (\Delta n + 1)}{2}$$
Here $\Delta n = 5 - 1 = 4$.
$$N = \frac{4 \times (4 + 1)}{2} = \frac{20}{2} = 10$$

Explanation:Using de Broglie relation $\lambda = \frac{h}{m v}$:
$$\lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.0 \times 10^6} = \frac{6.63 \times 10^{-34}}{1.82 \times 10^{-24}} = 3.64 \times 10^{-10}\text{ m} = 0.364\text{ nm}$$

Explanation:For a $4d$ orbital: principal quantum number $n = 4$ and azimuthal quantum number $l = 2$.
Radial nodes = $n - l - 1 = 4 - 2 - 1 = 1$.
Angular nodes = $l = 2$.
Thus, it has 1 radial node and 2 angular nodes.

Explanation:For a given principal quantum number $n$, the azimuthal quantum number $l$ can only take integer values from 0 to $n-1$. For $n=2$, the permissible values of $l$ are 0 and 1. Therefore, $l=2$ is strictly invalid.

Explanation:According to Einstein's photoelectric equation: $h\nu = \Phi + KE_{\text{max}}$.
Given photon energy $h\nu = 5.0\text{ eV}$ and work function $\Phi = 2.0\text{ eV}$:
$$KE_{\text{max}} = 5.0\text{ eV} - 2.0\text{ eV} = 3.0\text{ eV}$$

Explanation:From de Broglie relation, $\lambda = \frac{h}{p}$, so $\Delta x = \frac{h}{p}$.
Applying Heisenberg's principle $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$:
$$\left(\frac{h}{p}\right) \cdot \Delta p \ge \frac{h}{4\pi} \implies \frac{\Delta p}{p} \ge \frac{1}{4\pi}$$

Explanation:Chromium ($Z=24$) exhibits an anomalous configuration due to extra stability associated with half-filled $d$-subshells. One electron shifts from $4s$ to $3d$, resulting in $[\text{Ar}] 4s^1 3d^5$.

Explanation:Orbital angular momentum is given by $L = \sqrt{l(l+1)} \hbar = \sqrt{l(l+1)} \frac{h}{2\pi}$.
For a $p$ orbital, $l = 1$.
$$L = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$$

Explanation:For isoelectronic species (all having 10 electrons), ionic radius decreases as nuclear charge ($Z$) increases.
$\text{Mg}^{2+}$ has the highest nuclear charge ($Z = 12$), exerting the strongest electrostatic attraction on electrons, giving it the smallest ionic radius.

Explanation:For the first line of the Balmer series, $n_1 = 2$ and $n_2 = 3$.
$$\frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R_H \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36} R_H$$
$$\lambda = \frac{36}{5 \times 1.097 \times 10^7} = 6.563 \times 10^{-7}\text{ m} = 656.3\text{ nm}$$

Explanation:Number of neutrons $N = A - Z$.
For $^{13}_{6}\text{C}$: $N = 13 - 6 = 7$.
For $^{14}_{7}\text{N}$: $N = 14 - 7 = 7$.
Since both have 7 neutrons, they are isotones.

Explanation:The velocity of an electron in a Bohr orbit is $v_n \propto \frac{Z}{n}$, so $v_n = v_0 \left(\frac{Z}{n}\right)$.
For $\text{Li}^{2+}$ ($Z=3$) in the third orbit ($n=3$):
$$v = v_0 \left(\frac{3}{3}\right) = v_0$$

Explanation:The total number of orbitals in a principal energy level $n$ is given by $n^2$.
For $n = 3$, total orbitals $= 3^2 = 9$ (one $3s$, three $3p$, and five $3d$ orbitals). In a single-electron atom or isolated shell, all 9 orbitals are degenerate.

Explanation:The de Broglie wavelength is related to kinetic energy by $\lambda = \frac{h}{\sqrt{2m \cdot KE}}$.
If $KE' = 4 \cdot KE$, then:
$$\lambda' = \frac{h}{\sqrt{2m (4 \cdot KE)}} = \frac{1}{2} \frac{h}{\sqrt{2m \cdot KE}} = \frac{\lambda}{2}$$
Thus, the wavelength is halved.

Explanation:Frequency is proportional to energy change $\Delta E$.
For $n=2 \to n=1$: $\Delta E = 13.6 \left(1 - \frac{1}{4}\right) = 10.2\text{ eV}$.
For $n=3 \to n=2$: $\Delta E = 1.89\text{ eV}$.
For $n=4 \to n=3$: $\Delta E = 0.66\text{ eV}$.
For $n=5 \to n=4$: $\Delta E = 0.31\text{ eV}$.
The $n=2 \to n=1$ transition has the largest energy difference and highest frequency.

Explanation:For $n = 4$:
- Number of subshells $= n = 4$ (s, p, d, f)
- Total orbitals $= n^2 = 4^2 = 16$
- Max electrons $= 2n^2 = 2(16) = 32$

Explanation:The total number of orbitals in the $n=4$ shell is $n^2 = 4^2 = 16$.
According to Pauli's exclusion principle, each orbital can accommodate at most one electron with $m_s = +\frac{1}{2}$. Thus, maximum 16 electrons can share these quantum numbers.