The Hitomi satellite recently observed the Lyman alpha emission line ($n=2$ to $n=1$) of hydrogen-like iron ion (atomic number of iron is $26$) from the Perseus galaxy cluster. The wavelength of the line is closest to ............... $\mathring{A}$.

According to Bohr's theory,the radius of an electron in an orbit described by principal quantum number $n$ and atomic number $Z$ is proportional to:

As per the Bohr model,the minimum energy (in $eV$) required to remove an electron from the ground state of a doubly ionized $Li$ atom $(Z = 3)$ is:

An electron in the hydrogen atom excites from the $2^{nd}$ orbit to the $4^{th}$ orbit. The change in angular momentum of the electron is (Planck's constant $h = 6.64 \times 10^{-34} \ J \cdot s$).

Using the formula for the radius of the $n^{th}$ orbit $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,derive an expression for the total energy of an electron in the $n^{th}$ Bohr orbit.

$A$ proton is fired from a very large distance towards a nucleus with charge $Q = 120e$,where $e$ is the elementary charge. It reaches a distance of closest approach of $10 \ fm$. The de Broglie wavelength of the proton at its initial position is (in $fm$). (Given: mass of proton $m_p = (5/3) \times 10^{-27} \ kg$; $h/e = 4.2 \times 10^{-15} \ J \cdot s/C$; $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$; $1 \ fm = 10^{-15} \ m$)