If the ratio of the area of two orbits of an $H$ atom is $4 : 1$,then the ratio of the frequency of the $e^-$ in these two orbits is:

What is the energy of an electron in the stationary state corresponding to $n=2$?

The energy of the orbit of hydrogen is $E_n = \frac{1.31 \times 10^6}{n^2} \ J \ mol^{-1}$. If an electron transits from $n = 3$ to $n = 2$,find the frequency of the emitted radiation. (Note: $h = 6.6 \times 10^{-34} \ J \ s$,$N_A = 6.02 \times 10^{23} \ mol^{-1}$)

According to Bohr's model,the energy of an electron is given by $E = -\frac{2.176 \times 10^{-18}}{n^2} \, J \, atom^{-1}$. Calculate the minimum energy required to remove an electron from the $3^{rd}$ orbit of a $He^{+}$ ion,and also calculate the corresponding wavelength of the photon.

If in a hydrogen atom,an electron jumps from the $3^{rd}$ Lyman line to the $1^{st}$ Lyman line,then the obtained wavelength is:

Photon of the maximum frequency will be absorbed in the transition (for $H$ atom):