For HeX+\ce{He+}HeX+, 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=52.9a = 52.9a=52.9 pm; Rydberg constant, RH=2.2×10−18R_\mathrm{H} = 2.2\times10^{-18}RH=2.2×10−18 J; Planck's constant, h=6.6×10−34h = 6.6\times10^{-34}h=6.6×10−34 J s; Speed of light, c=3×108 m s−1c = 3\times10^8\ \mathrm{m\,s^{-1}}c=3×108 ms−1]
Official answer: 30
Which orbits. For HeX+\ce{He+}HeX+ (Z=2Z=2Z=2), rn=a n2Z=26.45 n2r_n=\frac{a\,n^2}{Z}=26.45\,n^2rn=Zan2=26.45n2 pm. So 105.8 pm is n=2n=2n=2 and 26.45 pm is n=1n=1n=1.
Photon energy.E=RHZ2(1−14)=2.2×10−18×4×34=6.6×10−18 J.E=R_HZ^2\left(1-\frac14\right)=2.2\times10^{-18}\times4\times\frac34=6.6\times10^{-18}\ \text{J}.E=RHZ2(1−41)=2.2×10−18×4×43=6.6×10−18 J.
Wavelength.λ=hcE=6.6×10−34×3×1086.6×10−18=3×10−8 m=30 nm.\lambda=\frac{hc}{E}=\frac{6.6\times10^{-34}\times3\times10^8}{6.6\times10^{-18}}=3\times10^{-8}\ \text{m}=30\ \text{nm}.λ=Ehc=6.6×10−186.6×10−34×3×108=3×10−8 m=30 nm.