For a hydrogen atom with $n = 1$,the radius of the Bohr orbit is approximately $0.530 \, \mathring{A}$. What will be the radius for the first excited state $(n = 2)$ in $\mathring{A}$?

Based on the equation $\Delta E = - 2.0 \times 10^{-18} \, J \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)$,the wavelength of the light that must be absorbed to excite a hydrogen electron from level $n = 1$ to level $n = 2$ will be: $(h = 6.625 \times 10^{-34} \, J \, s, c = 3 \times 10^8 \, m \, s^{-1})$

In a hydrogen atom,the energy difference between the states $n = 2$ and $n = 3$ is $E \ eV$. The ionization energy of the $H$-atom is: (in $E$)

In the ground state of $He^{+}$ ion,how much energy must be supplied to the electron so that the second line of the Balmer series is obtained in its spectrum? ................. $eV$

The ground and first excited state energies are $E_1$ and $E_2$ respectively. Which pair of species has the same energy? [Note that energy is indicated in the bracket].

The energy of a photon of wavelength $(\lambda) = 1 \, m$ is (Planck's constant $= 6.626 \times 10^{-34} \, J \, s$,speed of light $= 3 \times 10^8 \, ms^{-1}$)