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.
 

Answer 2

The radius of the nth orbit in a single-electron system is:
\(r=52.9\times\frac{n^2}{z}pm\)
Z is the atomic number.
For He⁺, Z is 2
Given, the initial radius 𝑟₂=105.8 pm and the final radius 𝑟₁=26.45 pm.
For 𝑟₂=105.8 pm, we get:
\(105.8=52.9\times\frac{n_2^2}{2}=\gt n_2=2\)
For 𝑟₁=26.45 pm, we get:
\(26.45=52.9\times\frac{n_1^2}{2}=\gt n_1=1\)
The energy difference is:
\(E=\frac{hc}{\lambda}=R_HZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\)

where h is Planck's constant, c is the speed of light, 𝑅𝐻 is the Rydberg constant, and 𝜆 is the photon's wavelength.
\(\frac{6.6\times10^{-34}J\ s\times3\times10^{8}m/s}{\lambda}=2.2\times10^{-18J\times2^2(\frac{1}{1^2})-\frac{1}{2^2})}\)
\(\lambda=30\times10^{-9}m,\ or\ 30nm\)

So, the answer is 30nm.