The kinetic energy of the electron in an orbit of radius $r$ in a hydrogen atom is ($e =$ electronic charge).

The energy required to remove one electron from a neutral helium atom is $24.6 \, eV$. The energy required to remove the remaining second electron (in $eV$) is .........

The figure shows a graph between $\ln \left| \frac{A_n}{A_1} \right|$ and $\ln |n|$, where $A_n$ is the area enclosed by the $n^{th}$ orbit in a hydrogen-like atom. The correct curve is

What is ionisation energy?

$A$ hydrogen atom,a deuteron atom,a $He^+$ ion,and a $Li^{++}$ ion each contain a single electron orbiting their nucleus. The wavelengths of the electromagnetic radiation emitted during the transition of the electron from the $n = 2$ orbit to the $n = 1$ orbit are found to be $\lambda_1, \lambda_2, \lambda_3$,and $\lambda_4$ respectively. Then:

$A$ free hydrogen atom after absorbing a photon of wavelength $\lambda_{a}$ gets excited from the state $n=1$ to the state $n=4$. Immediately after that,the electron jumps to $n=m$ state by emitting a photon of wavelength $\lambda_{e}$. Let the change in momentum of the atom due to the absorption and the emission be $\Delta p_{a}$ and $\Delta p_{e}$,respectively. If $\lambda_{a} / \lambda_{e} = 1/5$,which of the following options is/are correct?
[Use $hc = 1242 \text{ eV nm}$; $1 \text{ nm} = 10^{-9} \text{ m}$,$h$ and $c$ are Planck's constant and speed of light,respectively]
$(1)$ $\lambda_{e} = 418 \text{ nm}$
$(2)$ The ratio of kinetic energy of the electron in the state $n=m$ to the state $n=1$ is $1/4$
$(3)$ $m=2$
$(4)$ $\Delta p_{a} / \Delta p_{e} = 1/2$