In $(i)$ a hydrogen atom and $(ii)$ a singly ionized helium atom,an electron makes a transition from the same excited state $n$ to the ground state. The ratio of the wavelengths of the emitted photons in the two cases will be:

Which of the plots shown in the figure represents speed $(v)$ of the electron in a hydrogen atom as a function of the principal quantum number $(n)$?

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5 \text{ m s}^{-1}$. The radius of the orbit is $x \times 10^{-9} \text{ m}$. The value of $x$ is : [Take the mass of electron to be $9 \times 10^{-31} \text{ kg}$, charge of electron = $-1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$]

The radius of the smallest electron orbit in a hydrogen-like ion is $(0.51/4 \times 10^{-10}) \text{ m}$. This ion is:

The de-Broglie wavelength of the electron in the ground state of a hydrogen atom is

Hydrogen $(_1H^1)$,Deuterium $(_1H^2)$,singly ionized Helium $(_2He^4)^+$,and doubly ionized Lithium $(_3^6Li)^{++}$ all have one electron around the nucleus. Consider an electron transition from $n = 2$ to $n = 1$. If the wavelengths of emitted radiation are $\lambda_1, \lambda_2, \lambda_3$,and $\lambda_4$ respectively,then approximately which one of the following is correct?