Atomic Structure

The energy of an electron in a hydrogen atom is given by the formula:

$E_n = -\frac{R_{\mathrm{H}}}{n^2}$

We are told that the energy of the atom is equal to $-\frac{R_{\mathrm{H}}}{16}$. To find the principal quantum number $ n $, we compare the two expressions:

$-\frac{R_{\mathrm{H}}}{n^2} = -\frac{R_{\mathrm{H}}}{16}$

Canceling out $ -R_{\mathrm{H}} $ from both sides gives:

$\frac{1}{n^2} = \frac{1}{16}$

Therefore, $ n^2 = 16 \Rightarrow n = 4 $.

In a hydrogen atom, all orbitals that share the same principal quantum number $ n $ have the same energy. This means they are called degenerate. The total number of these degenerate orbitals is given by $ n^2 $.

So, for $ n = 4 $:

Degeneracy $ = n^2 = 4^2 = 16 $.

The de-Broglie wavelength is given by the formula:

$\lambda = \frac{h}{mV}$

According to the question, the wavelength is also given as:

$\lambda = 10 \sqrt{\frac{h}{m}}$

Now, equating the two expressions for $ \lambda $:

$\frac{h}{mV} = 10 \sqrt{\frac{h}{m}}$

Multiply both sides by $ mV $:

$h = 10 \, mV \sqrt{\frac{h}{m}}$

Divide both sides by 10:

$\frac{h}{10} = mV \sqrt{\frac{h}{m}}$

Square both sides to remove the square root:

$\left(\frac{h}{10}\right)^2 = (mV)^2 \times \frac{h}{m}$

Simplify the equation:

$\frac{h^2}{100} = m^2V^2 \frac{h}{m}$

$\frac{h}{100} = mV^2$

Now, solve for $ V $:

$V = \sqrt{\frac{h}{100m}}$

To find the ratio of speed to mass, $ \frac{V}{m} $:

$\frac{V}{m} = \frac{1}{m} \sqrt{\frac{h}{100m}} = \frac{\sqrt{h}}{10 \, m^{3/2}}$

The energy of a hydrogen atom is given by the equation:

$ E = -\frac{R_{H}}{9} $

which can also be expressed using the formula for energy levels:

$ E_{n} = -\frac{R_{\mathrm{H}}}{n^{2}} $

From these equations, we know:

$ n^{2} = 9 $

$ n = 3 $

For $ n = 3 $, the possible values of the azimuthal quantum number $ l $ are $ 0, 1, $ and $ 2 $, which correspond to the 3s, 3p, and 3d sub-shells, respectively.

The number of orbitals (which indicates the degeneracy) in each sub-shell is:

3s: 1 orbital

3p: 3 orbitals

3d: 5 orbitals

Adding these gives the total degeneracy:

$ 1 + 3 + 5 = 9 $

Thus, the degeneracy of the hydrogen atom with the given energy level is 9.

To find the value of the de Broglie wavelength in terms of its mass $ m $ and Planck's constant $ h $, we start by letting the wavelength of the given particle be $ x $. According to the problem, the velocity $ v $ is given by:

$ v = \frac{x}{100} $

Using the de Broglie wavelength formula, we have:

$ \lambda = \frac{h}{mv} $

Substituting the given velocity into the formula:

$ x = \frac{h}{m \left(\frac{x}{100}\right)} $

Simplifying this expression:

$ x = \frac{100h}{mx} $

Rearranging gives:

$ x^2 = 100 \frac{h}{m} $

Solving for $ x $, we find:

$ x = 10 \sqrt{\frac{h}{m}} $

Bohr's model of the hydrogen atom primarily describes the behavior of electrons in fixed orbits around the nucleus with quantized energy levels. Below is an analysis of each statement:

Option A: The position and velocity of the electrons in the orbit cannot be determined simultaneously.

This statement refers to the Heisenberg Uncertainty Principle and is not a part of Bohr's model. The Bohr model assumes electrons have well-defined orbits with known velocities and positions.

Option B: The electron in the orbit nearest the nucleus has the lowest energy.

This is indeed a part of Bohr's model. Bohr postulated that electrons occupy specific energy levels, and the closest orbit to the nucleus has the least energy.

Option C: Electrons revolve in different orbits around the nucleus.

This statement is a fundamental aspect of Bohr's model, where electrons move in discrete orbits or energy levels around the nucleus.

Option D: Energy of the electrons in the orbit is quantised.

This statement is also a key component of Bohr's model. Bohr proposed that the energy of electrons in these orbits is quantized, meaning electrons can only occupy certain energy levels.

Given the analysis, the statement that does not form a part of Bohr's model of hydrogen atom is:

Option A: The position and velocity of the electrons in the orbit cannot be determined simultaneously.

The hydrogen atom has the configuration 1s and 3s, 3p and 3d-orbitals will have same energy with respect to 1s-orbital.

The magnetic moment is given by $\mu = \sqrt {n(n + 2)} $ BM. (where, n = number of unpaired electron).

Electronic configuration of

$T{i^{2 + }}(22):1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}$

$\because$ Number of unpaired electron = 2

$\therefore$ $\mu = \sqrt {2(2 + 2)} = \sqrt 8 $

Energy for H-like atom, $(E) \propto {{{Z^2}} \over {{n^2}}}$

For hydrogen, Z = 1, n = 1

For lithium Z = 3, n = ?

$\therefore$ For same energy,

${E_H} = {E_{Li}} \Rightarrow {{Z_H^2} \over {n_H^2}} = {{Z_{Li}^2} \over {n_{Li}^2}}$

$\therefore$ $n_{Li}^2 = {{Z_{Li}^2 \times n_H^2} \over {Z_H^2}} \Rightarrow n_{Li}^2 = {{{3^2} \times 1} \over 1}$

$n_{Li}^2 = 3$

While angular momentum (nh/2$\pi$) is independent of atomic number.

$\therefore$ For 3s-orbital of lithium atom, energy and angular momentum is equal to that of 1s orbital of hydrogen atom.

The energy required for ionisation

H-atom is given by, $E = {{ - kZ{e^2}} \over {2r}}$

where, Z = atomic number

e = charge on proton/electron

r = radius of that electron

$\therefore$ $E \propto {1 \over r}$ or E $\propto$ r^$-$1

$\therefore$ n = $-$1

Rutherford model could not explain.

(i) Electronic structure of an atom.

(ii) Stability of an atom.

As we know,

$\nu = {1 \over \lambda } = {{{{(IE)}_H}{Z^2}} \over {hc}}x - \left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$

Given, n₁ = 2, n₂ = 5

${(IE)_H} = 13.6\,eV = 13.6 \times 1.6 \times {10^{ - 19}}\,J$

$\nu = {1 \over \lambda } = {{13.6 \times 1.6 \times {{10}^{ - 19}}\,J} \over {6.6 \times {{10}^{ - 34}}\,Js \times 3 \times {{10}^8}\,m{s^{ - 1}}}}\left[ {{1 \over {{2^2}}} - {1 \over {{5^2}}}} \right]$

$ = 2.2988 \times {10^6}\,{m^{ - 1}}$

$\therefore$ $\lambda = {1 \over \nu } \Rightarrow \lambda = 4.35 \times {10^{ - 7}}\,m = 435 \times {10^{ - 9}}\,m$

= 435 nm

If n = n, then I = 0, 1, 2 .... n

Thus, energy order is,

1s < 1p < 2s < 2p < 3s < 2d < 3p

Sc (21) will have electronic configuration as :

1s² 1p⁶ 2s² 2p⁶ 3s² 2d³