Problem 1

In 1894 Lord Rayleigh found that nitrogen prepared chemically (mean mass in a fixed vessel: 2.2990 g) was lighter than \“atmospheric nitrogen\” freed of

\ce{O2}

(2.3102 g), at 15.0 \°C and

1.013 \times 10^{5}

Pa; the difference was later attributed to argon. (a) Calculate the vessel volume and the mole fraction of Ar in atmospheric nitrogen. (b) Ramsay isolated helium from cleveite; it emits the D

_3

line at 587.7 nm. Calculate the photon energy and identify the corresponding transition in the He orbital diagram (answer: the highest-energy allowed transition, 3d \→ 2p) and the nuclear process that produces He in uranium minerals (

\alpha

-decay) and Ar in rocks (electron capture/positron decay of

\ce{^{40}K}

). (c) An unknown gas at 15.0 \°C and atmospheric pressure has density

0.850 \pm 0.005\ \text{kg m}^{-3}

; sound in it has

\lambda = 0.116

m at

f = 3520

Hz, with

v_s = f\lambda = \sqrt{\gamma RT/M}

. Find

M

and

\gamma

and identify the gas among HCl, HF, Ne, Ar.

Step 3 of 4: Helium D\₃ line and nuclear origins

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E = \dfrac{hc}{\lambda} = \dfrac{6.626 \times 10^{-34} \times 2.998 \times 10^{8}}{587.7 \times 10^{-9}} = 3.380 \times 10^{-19}\ \text{J}\;\;(\text{D}_3:\ 3d \rightarrow 2p);\quad \ce{^{238}U -> ^{234}Th + \alpha},\ \ \ce{^{40}K -> ^{40}Ar + \beta^+}(\varepsilon)

Analysis

The photon energy $3.380 \times 10^{-19}$ J matches the 3d \→ 2p transition, the largest allowed gap in the orbital diagram. Helium accumulates in uranium minerals through $\alpha$ -decay ( $\alpha$ particles are $\ce{He^{2+}}$ nuclei), and argon in rocks through $\ce{^{40}K}$ electron capture/positron emission.