The Correct Answer is Option (A)

The energy levels of a one-electron ion can be described by the formula:

$E_n=-\frac{Z^2}{n^2}\times E_0$

where $E_n$ is the energy of the nth level, Z is the atomic number (number of protons), n is the principal quantum number, and $E_0$is the ground state energy of the hydrogen atom (-13.6 eV).

For the ${\mathrm{He}}^{+}$ ion, the atomic number Z is 2 (since helium has 2 protons). We are looking for the energy of the first excited state, which corresponds to n = 2. Plugging these values into the formula, we get:

$E_2=-\frac{2^2}{2^2}\times (-13.6\mathrm{eV})=-13.6\mathrm{eV}$

So, the energy of the ${\mathrm{He}}^{+}$ion in its first excited state is$-13.6\mathrm{eV}$.

The Correct Answer is Option (A)

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck's constant). This can be expressed as:

$L=n\frac{h}{2\pi }$

where $n$ is the principal quantum number or the orbit number.

So, for the first orbit ($n=1$), the angular momentum $L_1$ is:

$L_1=1\times \frac{h}{2\pi }=\frac{h}{2\pi }$

And for the second orbit ($n=2$), the angular momentum $L_2$is:

$L_2=2\times \frac{h}{2\pi }=\frac{h}{\pi }$

The change in angular momentum when moving from the first to the second orbit is the difference between $L_2$ and $L_1$:

$\mathrm{\Delta }L=L_2-L_1=\frac{h}{\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }$

Since $\frac{h}{2\pi }$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.

Therefore, the correct answer is $L$.

The Correct Answer is Option (A)

According to Bohr's quantization of angular momentum, the angular momentum $L$ of a particle in a circular orbit is given by:

$L=n\hslash$

Where $n$ is an integer and $\hslash$ is the reduced Planck's constant. The angular momentum $L$ can also be expressed as:

$L=mvr$

Where $m$ is the mass of the particle,$v$is its linear velocity, and $r$is the radius of the orbit.

Now, we are given the potential energy $U=\frac{1}{2}m{\omega }^2{r}^2$. Since the particle is in a circular orbit, its centripetal force is provided by the gradient of the potential energy:

$m\frac{v^2}{r}=-\frac{\mathrm{d}U}{\mathrm{d}r}=-m{\omega }^{2}r$

We can simplify this equation to get the relation between $v$ and $r$:

$v^2={\omega }^2{r}^2$

Now, let's combine the equations for angular momentum and the relation between $v$ and $r$:

$n\hslash =mvr=m\sqrt{{\omega }^2{r}^2}r=m\omega r^2$

We can now solve for the radius $r$in terms of$n$:

$r^2=\frac{n\hslash }{m\omega }$

Taking the square root of both sides, we get:

$r\propto \sqrt{n}$

So, the radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$.

The Correct Answer is Option (A)

$r\propto \frac{n^2}{Z}$

$\begin{array}{rl}& \frac{r_{2nd}}{r_{3\mathrm{rd}}}={\left(\frac{n_2}{n_3}\right)}^2\\ \\ & \Rightarrow \frac{R}{r_{3rd}}={\left(\frac{2}{3}\right)}^2\\ \\ & \Rightarrow r_{3\mathrm{rd}}=\frac{9}{4}R\\ \\ & =2.25R\end{array}$

The Correct Answer is Option (C)

$v\alpha \frac{z}{n}$

$\frac{v_1}{v_2}=\left(\frac{n_2}{n_1}\right)$

$\Rightarrow \frac{3.6\times {10}^6}{v_2}=\frac{3}{7}$

$\Rightarrow v_2=\frac{7}{3}\times 3.6\times {10}^6$ m/s

$=8.4\times {10}^6$ m/s