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Natasha_Volkova [10]
3 years ago
6

A recent study found that electrons that have energies between 3.45 eV and 19.9 eV can cause breaks in a DNA molecule even thoug

h they do not ionize the molecule. If a single photon were to transfer its energy to a single electron, what range of light wavelengths could cause DNA breaks?minimum wavelength?
maximum wavelength?
Physics
1 answer:
vlada-n [284]3 years ago
3 0

Answer:

The Minimum wavelength is  \lambda_{min}= 382.2nm

The Maximum wavelength is \lambda_{max}= 624.2nm

Explanation:

From the question we are told that  

              The energy range is  E_r = 3.25eV \ and  \ 19.9eV

   Considering E = 19.9eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 19.9 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{max}}

Here h is the Planck's constant with value of  h= 6.625*10^{-34}J\cdot s

        c is the speed of light with value of  c = 3*10^8 m/s

Substituting values and making \lambda the subject of the formula

                       \lambda_{max} = \frac{hc}{E}

                         = \frac{6.625*10^{-34} * 3.0*10^{8}}{19.9*1.6*10^{-19}}

                         \lambda_{max}= 624.2nm

  Considering E = 3.25eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 3.25 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{min}}

Substituting values and making \lambda the subject of the formula

                       \lambda_{min} = \frac{hc}{E}

                           = \frac{6.625*10^{-34} * 3.0*10^{8}}{3.25*1.6*10^{-19}}

                           \lambda_{min}= 382.2nm

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The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

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Answer:

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Explanation:

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