69GP An oxygen atom at a particular s... [FREE SOLUTION]

Chapter 6: 69GP (page 138)

An 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.

Short Answer

Expert verified

The frequency after the sulfur substitution is\({f_{\rm{S}}} = 2.6 \times {10^{13}}\;{\rm{Hz}}\).

Step by step solution

Concept of relation of frequency with the mass

In order to predict the frequency after the sulfur substitution, the formula of frequency along with mass and spring constant will be utilized.

Given data

The frequency of oxygen atomis\({f_{\rm{O}}} = 3.7 \times {10^{13}}\;{\rm{Hz}}\).

The standard value of the mass of oxygen is \({m_{\rm{O}}} = 16\;{\rm{u}}\) and the mass of sulfur is \({m_{\rm{S}}} = 32\;{\rm{u}}\).

Calculation of frequency of sulfur atom

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\(2.6 \times {10^{13}}\;{\rm{Hz}}\).