The ionization energy of a hydrogen like Bohr atom is $4$ Rydbergs.
$(i)$ What is the wavelength of the radiation emitted when an electron jumps from the first excited state to the ground state?
$(ii)$ What is the radius of the first orbit for this atom?
A. $300\mathop A\limits^o ,\,2.65 \times {10^{ - 11}}m$
B. $600\mathop A\limits^o ,\,2.65 \times {10^{ - 11}}m$
C. $300\mathop A\limits^o ,\,4 \times {10^{ - 11}}m$
D. $600\mathop A\limits^o ,\,4 \times {10^{ - 11}}m$

Hint: As we know that the power needed to liberate the valence electron, which is the most loosely bonded electron in an isolated neutral gaseous atom or molecule. The energy ${E_1}$ of hydrogen has been provided. And the angular momentum, according to Bohr's theory, is an integral of Planck's constant divided by twice the value of $\pi $. Bohr's hypothesis only applied to one electron system, and in that case, the electron used the centripetal force to rotate around the nucleus.

Formula used:
When an electron transition from its first excited state to its ground state is given by:
${E_n} = \dfrac{{{Z^2}Rhc}}{{{n^2}}}$
And, for the hydrogen atom, we have the formula to determine the radius of the initial orbit:
$E = \dfrac{{hc}}{\lambda }$.

Complete step by step solution:
In the question, we have given that a hydrogen atom which is similar to a Bohr atom has an ionization energy of $4$ Rydbergs. The energy of photons produced when an electron transitions from the first excited state, $(n = 1)$, to the ground state is ${E_\infty } - {E_1}$, thus we have:
${E_\infty } - {E_1} = {Z^2}Rhc$
Based on the given information, we have:
${E_\infty } - {E_1} = {Z^2}Rhc = 4 \\$
$\Rightarrow {Z^2} = \dfrac{4}{{Rhc}} \\$
The value of $Rhc = 1\,rydberg$, so we can substitute in the above equation, then:
${Z^2} = \dfrac{4}{1} \\$
$\Rightarrow Z = \sqrt 4 \Rightarrow 2 \\$
In part $(i)$, The electron in a hydrogen atom is initially in the ground state $n = 1$,
${E_1} = \dfrac{{{Z^2}Rhc}}{{{{(1)}^2}}} \Rightarrow \dfrac{{{Z^2}Rhc}}{1}\,\,\,\,\,....(i)$
When an electron is in its first excited state, it has a $n = 2$,
${E_2} = \dfrac{{{Z^2}Rhc}}{{{{(2)}^2}}} \Rightarrow \dfrac{{{Z^2}Rhc}}{4}\,\,\,\,\,....(ii)$
Now, subtract $(ii)$ from $(i)$ to determine the energy required to excite the electron from $n = 1$ to $n = 2$, then:
${E_1} - {E_2} = \dfrac{{{Z^2}Rhc}}{1} - \dfrac{{{Z^2}Rhc}}{4} \\$
$\Rightarrow {E_1} - {E_2} = {Z^2}Rhc\left( {1 - \dfrac{1}{4}} \right) \\$
$\Rightarrow {E_1} - {E_2} = \dfrac{3}{4} \times 4 \\$
$\Rightarrow {E_1} - {E_2} = 3 \\$
Now, use the formula to determine the radius of the initial orbit for hydrogen atom,
$E = \dfrac{{hc}}{\lambda }$
If the wavelength of the radiation being emitted is $\lambda $, then we obtain:
$\Rightarrow \dfrac{{hc}}{\lambda } = 3 \\$
$\Rightarrow \lambda = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3 \times 2.2 \times {{10}^{ - 18}}}} \\$
$\Rightarrow \lambda = 300 \times {10^{ - 10}}m \\$
$\Rightarrow \lambda = 300\mathop A\limits^o \\$
Therefore, the wavelength of the radiation emitted is $300\mathop A\limits^o $.

In part $(ii)$, Let us consider the ${n^{th}}$bohr orbit, we have:
${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mZ{e^2}}}$
Where ${r_n}$ stands for the orbit's radius, $n$ for its fundamental quantum number, and $Z$ for its atomic number,
As we know that ${r_n} = \dfrac{{radius\,of\,first\,bohr \geqslant n}}{Z}$, then ${r_n} = \dfrac{{{r_0}}}{Z}$
Now, substitute the above values in the formula of radius of orbit, then we obtain:
${r_1} = \dfrac{{{{(1)}^2}{h^2}}}{{4{\pi ^2}m(2){e^2}}} \\$
$\Rightarrow {r_1} = \dfrac{1}{2}\left[ {\dfrac{{{h^2}}}{{4{\pi ^2}m{e^2}}}} \right] \\$
$\Rightarrow {r_1} = \dfrac{1}{2}\left( {0.529} \right) \\$
 $\therefore {r_1} = 0.2645\mathop A\limits^o \approx 2.65 \times {10^{ - 11}}m \\$
Therefore, the radius of the first orbit is $2.65 \times {10^{ - 11}}m$.

Thus, the correct option is A.

Note: When an electron in an atom absorbs energy, it moves from the lowest to the highest energy level, and when it moves from the highest to the lowest energy level, it releases energy in the form of light, which is referred to as emission of energy. The Bohr's radius formula used here applies to hydrogen that is isoelectronic, or has the same number of electrons as a hydrogen atom.