The correct answer is option (C)

The energy levels of a hydrogen atom are given by the formula :

$E_n=-\frac{13.6\mathrm{eV}}{n^2}$

where $E_n$ is the energy of the $n$-th energy level.

The ground state of hydrogen ($n=1$) has an energy of $-13.6\mathrm{eV}$ as mentioned, which means that it would take $+13.6\mathrm{eV}$ to ionize it (remove the electron completely) from this state, since ionization implies moving the electron to a state of zero energy.

The second excited state of hydrogen is when $n=3$ (as $n=1$ is the ground state and $n=2$ is the first excited state). Thus, the energy of the second excited state is :

$E_3=-\frac{13.6\mathrm{eV}}{3^2}=-\frac{13.6\mathrm{eV}}{9}=-1.51\mathrm{eV}$

Since ionization implies moving the electron from its current energy level to $0$ energy, the energy required to ionize the atom from this state is the absolute value of its current energy state. So, it will take $+1.51\mathrm{eV}$ to ionize a hydrogen atom from its second excited state.

So, the correct answer is Option C : $1.51\mathrm{eV}$.

The correct answer is option (A)

Given mvr $=1.5\frac{\mathrm{h}}{\pi }$

Compare with mvr $=\mathrm{n}\frac{\mathrm{h}}{2\pi }$

So $\frac{\mathrm{n}}{2}=1.5$ or $\mathrm{n}=3$

Now ${\mathrm{E}}_3=-\frac{13.6}{(3{)}^2}\mathrm{eV}\simeq -1.5\mathrm{eV}$

The correct answer is option (C)

Radius of n^th orbit in Hydrogen Atom

${\mathrm{r}}_{\mathrm{n}}=0.53\times \frac{{\mathrm{n}}^2}{\mathrm{Z}}$$\stackrel{o}{A}$

So, radius of third orbit

${\mathrm{r}}_3=0.53\times \frac{(3{)}^2}{(1)}$$\stackrel{o}{A}$ = 4.77 $\stackrel{o}{A}$

The correct answer is option (B)

Radius of Bohr's orbit depends on principal quantum number (n) as

$R\propto n^2$

Now, $\frac{R_1}{R_2}=\frac{{(2)}^2}{{(4)}^2}=\frac{1}{4}=0.25$

The correct answer is option (B)

Ratio of magnitude of Coulomb's electrostatic force to the gravitational force

$\frac{F_E}{F_G}=\frac{\left(\frac{K{q}_1{q}_2}{r^2}\right)}{\left(\frac{G{m}_1{m}_2}{r^2}\right)}=\frac{K{q}_1{q}_2}{G{m}_1{m}_2}$

$\Rightarrow 2.4\times {10}^{39}=\frac{K}{G}\times \frac{1.6\times {10}^{-19}\times 1.6\times {10}^{-19}}{9.11\times {10}^{-31}\times 1.67\times {10}^{-27}}$

$\Rightarrow 2.4\times {10}^{39}=\frac{K}{G}\times \frac{2.56}{15.21}\times {10}^{20}\Rightarrow \frac{K}{G}=14.26\times {10}^{19}$

$\Rightarrow \frac{K}{G}=1.426\times {10}^{20}$$\Rightarrow$ Ratio $\approx$ 10²⁰

The correct answer is option (C)

$E_n=\frac{E_0}{n^2}$, For first excited state$\Rightarrow$ n = 2

For second excited state $\Rightarrow$ n = 3

$\Rightarrow \frac{T_1}{T_2}=\frac{\frac{E_0}{4}}{\frac{E_0}{9}}=\frac{9}{4}$

The Correct Answer is Option (C)

Singly ionized neon has electron count more than one. Bohr's model is valid for atoms of single electron. So option (c) is not valid.

The Correct Answer is Option (A)

Apply Bohr's Atomic model for H-atom

KE = –T.E and PE = 2T.E

Given, T.E = –3.4 eV

$\therefore$ KE = +3.4 eV and PE = -6.8 eV

The Correct Answer is Option (B)

In a Bohr orbit of the hydrogen atom,

Kinetic energy = – (Total energy)

So, Kinetic energy : Total energy = 1 : –1

The Correct Answer is Option (D)

Energy of electron in He⁺ 3^rd orbit

E₃ = -13.6 $\times$ $\frac{4}{9}eV$

= -13.6 $\times$ $\frac{4}{9}$ $\times$ 1.6 $\times$ 10^-19 J

= - 9.7 $\times$ 10^-19 J

According to Bohr’s model,

Kinetic energy of electron in the 3^rd orbit = – E₃

$\therefore$ 9.7 $\times$ 10^-19 = $\frac{1}{2}{m}_e{v}^2$

v = $\sqrt{\frac{2\times 9.7\times {10}^{19}}{9.1\times {10}^{-31}}}$ = 1.46 $\times$ 10⁶ m/s

The Correct Answer is Option (D)

As T $\propto$ n³

$\therefore$ $\frac{T_1}{T_2}=\frac{8T_2}{T_2}$ = ${\left(\frac{n_1}{n_2}\right)}^3$

Hence, n₁ = 2n₂

The Correct Answer is Option (C)

E_n = $-\frac{13.6}{n^2}$

$\therefore$ E₁ = –13.6 eV

E₂ = –3.4 eV

E₃ = –1.5 eV

E₄ = –0.85 eV

$\therefore$ E₃ – E₂ = –1.5 – (–3.4) = 1.9 eV

E₄ – E₃ = –0.85 – (–1.5) = 0.65 eV

Obviously, difference of 11.1 eV is not possible.