Americas
Europe
69GP
Expert-verifiedAn oxygen atom at a particular site within a DNA molecule can be made to execute simple harmonic motion when illuminated by infrared light. The oxygen atom is bound with a spring-like chemical bond to a phosphorus atom, which is rigidly attached to the DNA backbone. The oscillation of the oxygen atom occurs with frequency \(f = 3.7 \times {10^{13}}\;{\rm{Hz}}\). If the oxygen atom at this site is chemically replaced with a sulfur atom, the spring constant of the bond is unchanged (sulfur is just below oxygen in the Periodic Table). Predict the frequency after the sulfur substitution.
The frequency after the sulfur substitution is \({f_{\rm{S}}} = 2.6 \times {10^{13}}\;{\rm{Hz}}\).
In order to predict the frequency after the sulfur substitution, the formula of frequency along with mass and spring constant will be utilized.
The frequency of oxygen atom is \({f_{\rm{O}}} = 3.7 \times {10^{13}}\;{\rm{Hz}}\).
The standard value of the mass of oxygen is \({m_{\rm{O}}} = 16\;{\rm{u}}\) and the mass of sulfur is \({m_{\rm{S}}} = 32\;{\rm{u}}\).
The relation of frequency and mass is calculated as:
\(\begin{aligned}{c}f &= \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{k}{m}} \\\frac{k}{{{\rm{4}}{{\rm{\pi }}^2}}} &= {f^2}m\\{f^2}m &= {\rm{constant}}\end{aligned}\)
Here, m is the mass and k is the spring constant.
The frequency of sulfur atom is calculated as:
\(\begin{aligned}{c}f_{\rm{O}}^2{m_{\rm{O}}} &= f_{\rm{S}}^2{m_{\rm{S}}}\\{f_{\rm{S}}} &= {f_{\rm{O}}}\sqrt {\frac{{{m_{\rm{O}}}}}{{{m_{\rm{S}}}}}} \\{f_{\rm{S}}} &= \left( {3.7 \times {{10}^{13}}\;{\rm{Hz}}} \right)\sqrt {\frac{{16\;{\rm{u}}}}{{32\;{\rm{u}}}}} \\{f_{\rm{S}}} &= 2.6 \times {10^{13}}\;{\rm{Hz}}\end{aligned}\)
Hence, the frequency of sulfur atom is \(2.6 \times {10^{13}}\;{\rm{Hz}}\).
94% of StudySmarter users get better grades.
Sign up for free