Atomic Structure

Number of electrons in lithium ion (Li⁺) is 2.

Number of electrons in helium is 2.

The spectrum of an atom depend on the number of electrons present in it.

Li$^+$ (Z = 3)

He (Z = 2)

(Both have same electrons = 2)

Bohr radius of hydrogen atom is first orbit in H-atom.

Bohr radius $({r_n}) = \left( {{{{n^2}} \over Z}} \right)\left( {{{{n^2}} \over {4{\pi ^2}m{e^2}}}} \right)$

${r_n} = {{{n^2}} \over Z} \times 0.529\,\mathop A\limits^o $

For first orbit $(n = 1),{r_1} = \left( {{1 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ ..... (i)

For second orbit $(n = 2),{r_2} = \left( {{4 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ ..... (ii)

For third orbit $(n = 3),{r_3} = \left( {{9 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ .... (iii)

Now divided (iii) by (i), we get

${r_3} = 9{r_1}$

The radius of 3rd orbit is 9 times the radius of 1st orbit.

The maximum number of possible electrons in a subshell with $n=3, l=2$ are 10 .

Because for, $n=3, l=2$, i.e. $d$-subshell has 10 electrons and 5 -orbital.

Hence, option (a) is the correct answer.

Given, uncertainty in position $(\Delta x)=1 \times 10^4 \mathrm{~m}$

uncertainty in velocity

$ (\Delta v)=0.627 \times 10^{-20} \mathrm{~ms}^{-1} $

Also,

$ \begin{aligned} & \because \quad \Delta x \cdot \Delta v=\frac{h}{4 \pi \cdot m} \\ & \text { } \begin{aligned} where, h & =\text { Plank's constant } \\ & =6.627 \times 10^{-34} \mathrm{Js} \\ m & =\text { to find (mass of particle) } \\ \therefore \quad m & =\frac{h}{\Delta x \cdot \Delta v \cdot 4 \pi} \\ & =\frac{6.627 \times 10^{-34}}{10^{-8} \times 6.627 \times 10^{-20} \times 4 \pi} \\ & =\frac{10^{-34} \times 10^{28}}{4 \pi} \text { in } \mathrm{kg} \\ \therefore \quad m & =\frac{10^{-34} \times 10^{28}}{4 \pi} \Rightarrow m=\frac{10^{-6}}{4 \pi} \mathrm{~kg} \end{aligned} \end{aligned} $

Hence, option (d) is the correct answer.

From de-Broglie's equation, $\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m \times \mathrm{KE}}}$

⇒ For an electrons, $\lambda \propto \frac{1}{\sqrt{\mathrm{KE}}} \Rightarrow \frac{\lambda_{16}}{\lambda_9}=\sqrt{\frac{9}{16}}=3: 4$

Regarding photoelectric effect, the statement of option-(c) is not correct. Because, the electrons are ejected from the metal surface as soon as the beam of light strikes to the surface.

This model could not explain the overall neutrality of the atom, because he proposed that the positive charge is spread over a sphere in which the electrons are embedded to make the atom as a whole neutral, but it was not consistant with the results. Hence, discarded.

In presence of electric and magnetic field, the cathode rays travel in straight line is not the result. Actually, they are deflected by electric and magnetic field.

When the rays are passed between two electrically charged plates, these are deflected towards positively charged plate.

In magnetic field, they are deflected at a right angle to both the field and to the direction of particle.

Given, $E=\frac{-R_{\mathrm{H}}}{16}$

We know that, $E=\frac{-R_{\mathrm{H}}}{n^2}=\frac{-R_{\mathrm{H}}}{16}, \therefore n=4$

Thus, degeneracy for $n=4$ will be,

Given, kinetic energy of $\mathrm{H}^{+}$ion $=1.68 \mathrm{eV}$

Mass of proton $=1.6726 \times 10^{-27} \mathrm{~kg}$

We calculate, the wavelength of $\mathrm{H}^{+}$ion as follows:

$ \lambda=\frac{h}{\sqrt{2 E_m}} $

Here, $E=$ kinetic energy, $m=$ mass of proton

$ \begin{aligned} h & =6.62 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} . \\ & =\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times E}} \\ & =\frac{0.286}{\sqrt{E(\mathrm{eV})}} \mathop {\rm{A}}\limits^{\rm{o}} =\frac{0.286}{\sqrt{1.68}}\mathop {\rm{A}}\limits^{\rm{o}} =0.220\mathop {\rm{A}}\limits^{\rm{o}} =0.022 \mathrm{~nm} \end{aligned} $