1) Frequency: ![3.29\cdot 10^{15}Hz](https://tex.z-dn.net/?f=3.29%5Ccdot%2010%5E%7B15%7DHz)
the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is
![E=2.18 aJ=2.18\cdot 10^{-18} J](https://tex.z-dn.net/?f=E%3D2.18%20aJ%3D2.18%5Ccdot%2010%5E%7B-18%7D%20J)
The energy of a photon is given by
![E=hf](https://tex.z-dn.net/?f=E%3Dhf)
where
is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:
![f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7BE%7D%7Bh%7D%3D%5Cfrac%7B2.18%5Ccdot%2010%5E%7B-18%7D%20J%7D%7B6.63%5Ccdot%2010%5E%7B-34%7D%20Js%7D%3D3.29%5Ccdot%2010%5E%7B15%7D%20Hz)
2) Wavelength: 91.2 nm
The wavelength of the photon can be found from its frequency, by using the following relationship:
![\lambda=\frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bc%7D%7Bf%7D)
where
is the speed of light and f is the frequency. Substituting the frequency, we find
![\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B3%5Ccdot%2010%5E8%20m%2Fs%7D%7B3.29%5Ccdot%2010%5E%7B15%7DHz%7D%3D9.12%5Ccdot%2010%5E%7B-8%7D%20m%3D91.2%20nm)
To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,
Frequency ![= f = 562Hz](https://tex.z-dn.net/?f=%3D%20f%20%3D%20562Hz)
Speed of sound in air ![= v = 331m/s](https://tex.z-dn.net/?f=%3D%20v%20%3D%20331m%2Fs)
The definition of wavelength is,
![\lambda = \frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bv%7D%7Bf%7D)
Here,
v = Velocity
f = Frequency
Replacing,
![\lambda = \frac{331m/s}{562Hz}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B331m%2Fs%7D%7B562Hz%7D)
![\lambda = 0.589m](https://tex.z-dn.net/?f=%5Clambda%20%3D%200.589m)
Therefore the wavelength of that tone in air at standard conditions is 0.589m
Eric is writing about the cell wall.
Chloroplast is for photosynthesis.
Mitochondria releases energy from respiration.
Nucleus controls the activities of the cell.
But the cell wall supports and gives structure to the cell.