Question

A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student prepares a set of CoCl2(aq) solutions of known concentration. Then the student uses a spectrophotometer to determine the absorbance of each of the standard solutions at a wavelength of 510nm and constructs a standard curve. Finally, the student determines the absorbance of the sample of unknown concentration. A wavelength of 510nm corresponds to an approximate frequency of 6×1014s−1. What is the approximate energy of one photon of this light? 9×1047J.

Answer 1

The approximate energy of one photon of this light  is ; $4*10^{-9} J$

Given that :

wavelength of 510nm ≈ frequency of $6*10^{14} s^{-1}$

Energy of photon ( E ) = hv = $\frac{hc}{\beta }$   ---- ( 1 )

where;

h = Planck's constant = $6.63 * 10^{-34}Js$

c = $3*10^8 m/s$

β = light wavelength = $510 * 10^{-9} m$

v = light frequency

Insert given values into equation ( 1 )

E = $4*10^{-9} J$

Hence we can conclude that approximate energy of one photon of this light is  $4*10^{-9} J$

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Answer 2

The energy of the photon will be $\rm \bold {E = 4 \times 10^-9 J}$

The energy of photon,

$\rm \bold{ E = \frac{hc}{\lambda} }$

where,

h- plank scale = $\rm \bold { 6.63 \times 10^-34 Js}$

c- $\rm \bold { 3 \times 10^8 m/s}$

$\rm \bold{ \lambda} }$-  wavelength = 510nm

Put the values in the formula

$\rm \bold {E = 4 \times 10^-9 J}$

Hence, we can conclude that the energy of the photon will be $\rm \bold {E = 4 \times 10^-9 J}$

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