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Year 1 of the natural sciences course of the University of Cambridge

Chemistry course of the year 1 of the natural sciences course of the University of Cambridge

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge

2025

An emitted <photon> has positive <energy>
$$
E_\gamma=\frac{hc}{\lambda}
=\frac{(6.626\times10^{-34})(2.998\times10^8)}{18.2\times10^{-9}}
=1.09147\times10^{-17}\,\mathrm J.
$$
For <spontaneous emission> into $n_f=2$, the <conservation of energy> equation is
$$
E_\gamma=E_{n_i}-E_2=36R_H\left(\frac14-\frac1{n_i^2}\right).
$$
Consequently
$$
\frac1{n_i^2}=\frac14-\frac{1.09147\times10^{-17}}{7.848\times10^{-17}}=0.110924,
\qquad n_i=3.0025.
$$
The <principal quantum number> must be an integer, so \b[the transition is $\boxed{n_i=3\longrightarrow n_f=2}$.] Its calculated <wavelength> is approximately $18.22\,\mathrm{nm}$, agreeing with the rounded observation.

Solution

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 2 / a / iii / Solution

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 2 / a / iii / Solution

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