
Greenlight has a wavelength of $5200\,{{\rm A}^\circ }$. How do you calculate the energy of one photon of green light?
Answer
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Hint: In this question, we are asked to find the energy of a photon of green light and we are provided with the wavelength of the green light. We will be using S.I. units in this question. We need to convert the wavelength value from angstroms to meters using the conversion factor.
Formula used:
We are performing this question in S.I. units.
$E = \,\dfrac{{hc}}{\lambda }$
Where E is the energy of one photon (in joules)
h is Planck’s constant whose value is $6.626 \times {10^{ - 34}}\,J\,s$
c is the speed of the light in vacuum whose value is $3 \times {10^8}\,m\,{s^{ - 1}}$
$\lambda $is the wavelength of light in units of meters
Complete solution:
We are provided with the wavelength of green light equal to $5200\,{{\rm A}^\circ }$
Converting this in S.I. units,
We know that, $1\,{A^\circ } = {10^{ - 10}}m$
$\lambda = 5200{A^\circ } = 5200 \times {10^{ - 10}} = 5.2 \times {10^{ - 7}}m$
Using the formula, calculating energy from the value of wavelength
$E = \,\dfrac{{hc}}{\lambda }$
h=$6.626 \times {10^{ - 34}}\,J\,s$
c= $3 \times {10^8}\,m\,{s^{ - 1}}$
$\lambda = 5200{A^\circ } = 5200 \times {10^{ - 10}} = 5.2 \times {10^{ - 7}}m$
Substituting all the values in the given formula, we get
$E = \dfrac{{6.626 \times {{10}^{ - 34}}J\,s\, \times 3 \times {{10}^8}m{s^{ - 1}}}}{{5.2 \times {{10}^{ - 7}}m}}$
= $3.82 \times {10^{ - 19\,}}J$(rounded to 2 decimal places)
Which is the required answer.
Note: We need to take special care of the units while solving these types of questions. It’s better to perform such calculations in S.I. units. It’s better to learn all the values of the constants as we can see these are not mentioned in the question itself. We can see that wavelength and energy has an inverse relationship. More the value of wavelength, less will be the value of energy and vice versa.
Formula used:
We are performing this question in S.I. units.
$E = \,\dfrac{{hc}}{\lambda }$
Where E is the energy of one photon (in joules)
h is Planck’s constant whose value is $6.626 \times {10^{ - 34}}\,J\,s$
c is the speed of the light in vacuum whose value is $3 \times {10^8}\,m\,{s^{ - 1}}$
$\lambda $is the wavelength of light in units of meters
Complete solution:
We are provided with the wavelength of green light equal to $5200\,{{\rm A}^\circ }$
Converting this in S.I. units,
We know that, $1\,{A^\circ } = {10^{ - 10}}m$
$\lambda = 5200{A^\circ } = 5200 \times {10^{ - 10}} = 5.2 \times {10^{ - 7}}m$
Using the formula, calculating energy from the value of wavelength
$E = \,\dfrac{{hc}}{\lambda }$
h=$6.626 \times {10^{ - 34}}\,J\,s$
c= $3 \times {10^8}\,m\,{s^{ - 1}}$
$\lambda = 5200{A^\circ } = 5200 \times {10^{ - 10}} = 5.2 \times {10^{ - 7}}m$
Substituting all the values in the given formula, we get
$E = \dfrac{{6.626 \times {{10}^{ - 34}}J\,s\, \times 3 \times {{10}^8}m{s^{ - 1}}}}{{5.2 \times {{10}^{ - 7}}m}}$
= $3.82 \times {10^{ - 19\,}}J$(rounded to 2 decimal places)
Which is the required answer.
Note: We need to take special care of the units while solving these types of questions. It’s better to perform such calculations in S.I. units. It’s better to learn all the values of the constants as we can see these are not mentioned in the question itself. We can see that wavelength and energy has an inverse relationship. More the value of wavelength, less will be the value of energy and vice versa.
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