Question

For He+, a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm. The wavelength (in nm) of the emitted photon during the transition is ___.
[Use: Bohr radius, a = 52.9 pm, Rydberg constant, 𝑅H = 2.2 × 10−18 J, Planck’s constant, h = 6.6 × 10−34 J s, Speed of light, c = 3 × 108 m s−1]

Answer 1

$r=52.9\times\frac{n^2}{z}\,\,pm$
$\therefore\,105.8=\frac{52.9\times n^2}{2}\,\,\,\,\,\,\,\,\therefore n_2=2$
and $26.45=52.9\times\frac{n^2}{2}\,\,\,\,\,\,\,\,\therefore n_1=1$
$\because$ $\Delta E=R_HhC\times z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{hc}{\lambda}=R_HhC\times z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{6.6\times10^{-34}\times3\times10^8}{\lambda}=2.2\times10^{-18}\times4\times\frac{3}{4}$
$\therefore$ $\lambda=300Å$
$\therefore\,\lambda=30\,nm$
The wavelength of the emitted photon is 30 nm.