Respuesta :
1 mol ≡ 6.023 x 10^23 molecules are produced using 467000 j of energy,
then production of 1 molecule will required energy = [tex] \frac{467000}{6.023X10^2^3} [/tex] = 7.7536 x 10^(-19) j
Thus, mininum energy of a photon that can free a H-atom from water is 7.7536 x 10^(-19) j.
Now, we know that E = hv
where, h = planck's constant = 6.63 10^(34) J.s
v = frequency of photon
∴ v = [tex] \frac{E}{h} [/tex] = [tex] \frac{7.7536 X 10^(^-^1^9^)}{6.63X10^(^-^3^4^)} [/tex] = 1.1702 X 10^(15) Hz
We also know that, wavelength (∧) = [tex] \frac{c}{v} [/tex]
= [tex] \frac{3X10^8}{1.1702X10^(^1^5^)} [/tex]
= 2.56 X 10^(-7) m
then production of 1 molecule will required energy = [tex] \frac{467000}{6.023X10^2^3} [/tex] = 7.7536 x 10^(-19) j
Thus, mininum energy of a photon that can free a H-atom from water is 7.7536 x 10^(-19) j.
Now, we know that E = hv
where, h = planck's constant = 6.63 10^(34) J.s
v = frequency of photon
∴ v = [tex] \frac{E}{h} [/tex] = [tex] \frac{7.7536 X 10^(^-^1^9^)}{6.63X10^(^-^3^4^)} [/tex] = 1.1702 X 10^(15) Hz
We also know that, wavelength (∧) = [tex] \frac{c}{v} [/tex]
= [tex] \frac{3X10^8}{1.1702X10^(^1^5^)} [/tex]
= 2.56 X 10^(-7) m
The minimum energy of a photon that can produce one hydrogen atom from water is [tex]\boxed{7.755 \times {{10}^{ - 19}}{\text{ J}}}[/tex]
The frequency of the photon is [tex]\boxed{1.17 \times {{10}^{15}}{\text{ }}{{\text{s}}^{ - 1}}}[/tex].
The wavelength of the photon is [tex]\boxed{2.563 \times {{10}^{ - 7}}{\text{ m}}}[/tex].
Further Explanation:
Frequency[tex]\left( \nu \right)[/tex] is defined as number of times n event occurs in unit time. It is generally applied to waves including light, sound, and radio waves. It is denoted by [tex]{\nu }}[/tex] and its SI unit is Hertz (Hz).
Wavelength is the characteristic property of a wave. It is defined as the distance between two successive crests or troughs. A crest is that point where there is maximum displacement of the medium whereas trough is a point that has minimum displacement of the medium. It is represented by [tex]\lambda[/tex] and its SI unit is meter (m).
First, the energy required to produce one hydrogen atom in joule can be calculated as follows:
[tex]E\left({\text{J}}\right)=\frac{{E\left( {{\text{kJ}}}\right)\times\left({\frac{{{{10}^3}{\text{ J}}}}{{{\text{kJ}}}}}\right)}}{{{{\text{N}}_{\text{A}}}}}[/tex] …… (1)
Here, [tex]{{\text{N}}_{\text{A}}}[/tex] is an Avogadro's number and has a value [tex]6.022 \times {10^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}[/tex].
Substitute [tex]467{\text{ kJ/mol}}[/tex] for [tex]E\left( {{\text{kJ}}} \right)[/tex] and [tex]6.022 \times {10^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}[/tex] for [tex]{{\text{N}}_{\text{A}}}[/tex] in equation (1).
[tex]\begin{aligned}E\left( {\text{J}} \right)&=\frac{{\left({467{\text{ kJ/mol}}} \right) \times \left( {\frac{{{{10}^3}{\text{ J}}}}{{{\text{kJ}}}}}\right)}}{{6.022\times {{10}^{23}}{\text{ mo}}{{\text{l}}^{-1}}}}\\&=7.755\times{10^{-19}}{\text{ J}}\\\end{aligned}[/tex]
Thus the energy of a photon that can free one atom of hydrogen is [tex]7.755 \times {10^{ - 19}}{\text{ J}}[/tex].
The expression of frequency and energy is as follows:
[tex]E=hv[/tex] …… (2)
Here, [tex]v[/tex] is a frequency of photon and h is a Plank’s constant and has a value [tex]\left({6.626\times{{10}^{-34}}{\text{ Js}}}\right)[/tex].
Rearrange equation (2) to calculate the frequency of the photon as follows:
[tex]v=\frac{E}{h}[/tex] …… (3)
Substitute [tex]6.626\times{10^{-34}}{\text{ J}}\cdot{\text{s}}[/tex] for h and [tex]7.755 \times {10^{-19}}{\text{ J}}[/tex] for E in equation (3).
[tex]\begin{aligned}v&=\frac{{7.755\times{{10}^{-19}}{\text{ J}}}}{{6.626\times{{10}^{-34}}{\text{ Js}}}}\\&=1.17\times{10^{15}}{\text{}}{{\text{s}}^{-1}}\\\end{aligned}[/tex]
Thus the frequency of photon is [tex]1.17\times{10^{15}}{\text{}}{{\text{s}}^{-1}}[/tex].
The expression to calculate the wavelength from energy of the photon is as follows:
[tex]E = \frac{{h{\text{c}}}}{{\lambda }}}[/tex] …… (4)
Here [tex]{\lambda }}[/tex] is a wavelength of a photon and c is a speed of light.
Rearrange equation (4) to calculate wavelength of the photon as follows:
[tex]{\lambda }}=\frac{{h{\text{c}}}}{E}[/tex] …… (5)
Substitute [tex]6.626 \times {10^{ - 34}}{\text{ J}} \cdot {\text{s}}[/tex] for h, [tex]3.0 \times {10^8}{\text{ m/s}}[/tex] for c and [tex]7.755 \times {10^{ - 19}}{\text{ J}}[/tex] for E in equation (5).
[tex]\begin{aligned}{\lambda}}=\frac{{\left({6.626\times {{10}^{-34}}{\text{ J}}\cdot {\text{s}}} \right)\left( {3.0 \times {{10}^8}{\text{ m/s}}}\right)}}{{\left({7.755\times {{10}^{ - 19}}{\text{ J}}} \right)}}\\=2.563\times {10^{-7}}{\text{m}}\\\end{aligned}[/tex]
Hence wavelength of the photon is equal to [tex]2.563 \times {10^{ - 7}}{\text{ m}}[/tex].
Learn more:
1. Ranking of photons according to the wavelength of transition: https://brainly.com/question/2055545
2. Calculate de Broglie wavelength of golf ball: https://brainly.com/question/7047430
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Structure of atom
Keywords: Hydrogen atom, cheapest source of h2, 467 kj, 1 mol of h atom, frequency wavelength.
