The magnitude of acceleration of the electron in the $n$th orbit of a hydrogen atom is $a_{H}$ and that of a singly ionised helium atom is $a_{He}$. The ratio of $a_{H} : a_{He}$ is

$A$ hydrogen atom is in its $n^{\text{th}}$ energy state. If the de-Broglie wavelength of the electron is $\lambda$,then:

What is the angular momentum of an electron in the $n=5$ orbit of a $Be^{+3}$ ion?

The de-Broglie wavelength of an electron in the $3^{rd}$ orbit of a $He^{+1}$ ion is approximately (in $\mathring{A}$):

In a hydrogen atom,the electron is moving around the nucleus with a velocity of $2.18 \times 10^{6} \; m/s$ in an orbit of radius $0.528 \; \mathring{A}$. The acceleration of the electron is:

For a certain hypothetical one-electron atom,the wavelength (in $\mathring{A}$) for the spectral lines for transition from $n = p$ to $n = 1$ is given by $\lambda = \frac{1500p^2}{p^2 - 1}$ (where $p > 1$). The ionization potential of this element must be .....$V$ (Take $hc = 12420\ eV\cdot\mathring{A}$).