Physics M.C.Q [1M]

4. Doubly ionized lithium.

Explanation:

The wavelength corresponding the transition from n₂ to n₁ is given by,

$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)$

Here,

R = Rydberg constant.

Z = Atomic number of the ion.

From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.

Therefore, the wavelength corresponding to n = 2 to n = 1 will be minimum in doubly ionized lithium ion because for lithium, Z = 3.

1. 1.05 × 10^-34Js

Explanation:

Let after absorption of energy, the hydrogen atom goes to the nth excited state.

Therefore, the energy absorbed can be written as,

$10.2=13.6\times\Big(\frac{1}{1^2}-\frac{1}{\text{n}^2}\Big)$

$\Rightarrow\frac{10.2}{13.6}=1-\frac{1}{\text{n}^2}$

$\Rightarrow\frac{1}{\text{n}^2}=\frac{13.6-10.2}{13.6}$

$\Rightarrow\frac{1}{\text{n}^2}=\frac{3.4}{13.6}$

$\Rightarrow\text{n}^2=4$

$\Rightarrow\text{n}=2$

The orbital angular momentum of the electron in the n^th state is given by,

$\text{L}_\text{n}=\frac{\text{nh}}{2\pi}$

Change in the angular momentum, $\Delta\text{L}=\frac{2\text{h}}{2\pi}-\frac{\text{h}}{2\pi}=\frac{\text{h}}{2\pi}$

$\therefore\Delta\text{T}=1.05\times10^{-34}\text{Js}$

4. Doubly ionized lithium.

Explanation:

For a hydrogen-like ion with Z protons in the nucleus, the radius of the nth state is given by,

$\text{r}_\text{n}=\frac{\text{n}^2\text{a}_0}{\text{Z}}$

Here,

a₀ = 0.53 pm

For lithium, Z = 3 Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

3. $\frac{\text{h}}{2\pi}$

Explanation:

According to Bohr's atomic theory, the orbital angular momentum of an electron is an integral multiplt of $\frac{\text{h}}{2\pi}$

$\therefore\ \text{L}_\text{n}=\frac{\text{nh}}{2\pi}$

Here,

n = Principal quantum number

The minimum of n is 1

Thus, the minimum value of the orbit angular momentum of the electron in a hydrogen is given by $\text{L}=\frac{\text{h}}{2\pi}$ 

1. Will pass through the origin.
2. Will be a straight line with slope 4.

Explanation:

$\text{r}_\text{n}=\text{n}^2\text{a}_0$

Area of the nth orbit is given by,

$\text{A}_\text{n}=\pi\text{r}^2_\text{n}=\pi\text{n}^4\text{a}^2_0$

$\text{A}_1=\pi\text{a}^{2}_0$

$\text{ln}\Big(\frac{\text{A}_\text{n}}{\text{A}_1}\Big)=\text{ln}\Big(\frac{\pi\text{n}^4\text{a}_0^2}{\pi\text{a}^2_0}\Big)$

$\text{ln}\Big(\frac{\text{A}_\text{n}}{\text{A}_1}\Big)=4\text{ln n}\ ...(\text{i})$

From the above expression, the graph of in $\Big(\frac{\text{A}_\text{n}}{\text{A}_1}\Big)$ against ln(n) will be a straight line passing through the origin and having slope 4.

1. vn.
2. Er.

Explanation:

Relations for energy, radius of the orbit and its velocity are given by,

$\text{E}=-\frac{\text{mZ}^2e^4}{8\in_0^2\text{h}^2\text{n}^2}$

$\text{r}=\frac{\in_0\text{h}^2\text{n}^2}{\pi\text{mZe}^2}$

$\text{v}=\frac{\text{Ze}^2}{2\in_0\text{hn}}$

Where,

Z: The atomic number of hydrogen like atom.

e: Electric charge.

h: Plank constant.

m: Mass of electron.

n: Principal quantam number of the electron.

$\in_0:$ Permittivity of vacuum

From these relations, we can see that the products independent of n are vn, Er.

3. 10^-42Nm

Explanation:

The angular momentum of the electron for the n^th state is given by,

$\text{L}_\text{n}=\frac{\text{nh}}{2\pi}$

Angular momentum of the electron for $\text{n}=3,\ \text{L}_\text{i}=\frac{3\text{h}}{2\pi}$

Angular momentum of the electron for $\text{n}=2,\ \text{L}_\text{f}=\frac{2\text{h}}{2\pi}$

The torque is the time rate of change of the angular momentum.

Torque, $\tau=\frac{\text{L}_\text{f}-\text{L}_\text{i}}{\text{t}}$

$=\frac{\big(\frac{2\text{h}}{2\pi}\big)-\big(\frac{3\text{h}}{2\pi}\big)}{10^{-8}}$

$=\frac{-\big(\frac{\text{h}}{2\pi}\big)}{10^{-8}}$

$=\frac{-10^{-34}}{10^{-8}}$ $\Big[\because\frac{\text{h}}{2\pi}\approx10^{-34}\text{J}-\text{s}\Big]$

$=-10^{-42}\text{N}-\text{m}$

The magnitude of the torque is 10^-42N-m.

4. Orbital angular momentum of the electron.

Explanation:

According to Bohr's atomic theory, the orbital angular momentum of an electron in a one-electron system is given by,

$\text{L}_\text{n}=\frac{\text{nh}}{2\pi}$

Here,

n = Principal quantum number.

The angular momentum is independent of the atomic number of the one-electron system. Therefore, it is same for all hydrogen-like atoms and ions in their ground states. The other parameters given here are dependent on the atomic number of the hydrogen-like atom or ion taken.

Explanation:

The speed (v) of electron can be expressed as,

$\nu=\frac{\text{Ze}^2}{2\in_0\text{hn}}\ ....(\text{i})$

Here,

Z = Number of protons in the nucleus.

e = Magnitude of charge on electron charge.

n = Principal quantum number.

h = Planck's constant.

It can be observed from equation (i) that the velocity of electron is inversely proportional to the principal quantum number (n).

Therefore, the graph between them must be a rectangular hyperbola. The correct curve is (c).

2. u_A > u_B

Explanation:

The ionisation energy of a hydrogen like ion of atomic number Z is given by,

$\text{V}=(13.6\text{eV})\times\text{Z}^2$

Thus, the atomic number of ion A is greater than that of B(Z_A > Z_B).

The radius of the orbit is inversely proportional to the atomic number of the ion.

$\therefore\ \text{r}_\text{A}>\text{r}_\text{B}$

The speed of electron is directly proportional to the atomic number.

Therefore, the speed of the electron in the orbit of A will be more than that in B.

Thus, u_A > u_B is correct.

The total energy of the atom is given by,

$\text{E}=-\frac{\text{mZ}^2\text{e}^2}{8\in_0\text{h}^2\text{n}^2}$

As the energy is directly proportional to Z², the energy of A will be less than that of B, i.e. E_A < E_B.

The orbital angular momentum of the electron is independent of the atomic number. Therefore, the relation L_A > L_B is invalid.

4. n = 2 to n = 1

Explanation:

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by,

$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}_1}-\frac{1}{\text{n}_2}\Big)$

Here, R is the Rydberg constant.

For the transition from n = 5 to n = 4, the wavelength is given by,

$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{4^2}-\frac{1}{5^2}\Big)$

$\lambda=\frac{400}{9\text{RZ}^2}$

For the transition from n = 4 to n = 3, the wavelength is given by,

$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{3^2}-\frac{1}{4^2}\Big)$

$\lambda=\frac{144}{7\text{RZ}^2}$

For the transition from n = 3 to n = 2, the wavelength is given by,

$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)$

$\lambda=\frac{36}{5\text{RZ}^2}$

For the transition form n = 2 to n = 1, the wavelength is given by,

$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$

$\lambda=\frac{2}{\text{RZ}^2}$

From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from n = 2 to n = 1 will be minimum.

4. Li⁺⁺

Explanation:

The radius of the n^th orbit in one electron system is given by,

$\text{r}_\text{n}=\frac{\text{n}^2\text{a}_0}{\text{Z}}$

Here, a₀ = 53 pm

For the shortest orbit, n = 1

For hydrogen, Z = 1

$\therefore$ Radius of the first state of hydrogen atom = 53 pm

For deuterium, Z = 1

$\therefore$ Radius of the first state of deuterium atom = 53 pm

For He⁺, Z = 2

$\therefore$ Radius of He⁺ atom $=\frac{53}{2}\text{ pm}=26.5\text{ pm}$

For Li⁺⁺, Z = 3

$\therefore$ Radius of Li⁺⁺ atom $=\frac{53}{3}\text{ pm}=17.66\approx18\text{ pm}$ 

The given one-electron system having radius of the shortest orbit to be 18 pm may be Li⁺⁺

3. He⁺

Explanation:

The total energy of a hydrogen-like ion, having Z protons in its nucleus, is given by,

$\text{E}=-\frac{13.6\text{Z}^2}{\text{n}^2}\text{eV}$

Here, n = Principal quantum number.

For ground state, n = 1

$\therefore$ Total energy, E = -13.6Z²eV

For hydrogen, Z = 1

$\therefore$ Total energy, E = -13.6eV

For deuterium, Z = 1

$\therefore$ Total energy, e = -13.6eV

For He⁺, Z = 2

$\therefore$ Total energy E = -13.6 × 2² = -54.4eV

For Li⁺⁺,

Z = 3

$\therefore$ Total energy, E = -13.6 × 3² = -122.4eV

Hence, the ion having an energy of -54.4eV in its ground state may be He⁺

2. Increases.

Explanation:

The electric potential energy of hydrogen atom with electron at the n^th state is given by,

$\text{V}=-\frac{2\times13.6}{\text{n}^2}$

As the value of n increases, the potential energy of the hydrogen atom also increases, i.e. the atom becomes less bound as n increases.