Chemistry, 21.07.2019 23:00 trinidymwilga

In a future hydrogen-fuel economy, the cheapest source of h2 will certainly be water. it takes 467 kj to produce 1 mol of h atoms from water. what is the frequency, wavelength, and minimum energy of a photon that can free an h atom from water? enter your answers in scientific notation.

Answers: 1

Show answers

Answers

1 mol ≡ 6.023 x 10^23 molecules are produced using 467000 j of energy,
then production of 1 molecule will required energy = $\frac{467000}{6.023X10^2^3}$ = 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 = $\frac{E}{h}$ = $\frac{7.7536 X 10^(^-^1^9^)}{6.63X10^(^-^3^4^)}$ = 1.1702 X 10^(15) Hz

We also know that, wavelength (∧) = $\frac{c}{v}$
= $\frac{3X10^8}{1.1702X10^(^1^5^)}$
= 2.56 X 10^(-7) m

Answer from: hahahwha

The minimum energy of a photon that can produce one hydrogen atom from water is $\boxed{7.755 \times {{10}^{ - 19}}{\text{ J}}}$

The frequency of the photon is $\boxed{1.17 \times {{10}^{15}}{\text{ }}{{\text{s}}^{ - 1}}}$ .

The wavelength of the photon is $\boxed{2.563 \times {{10}^{ - 7}}{\text{ m}}}$ .

Further Explanation:

Frequency $\left( \nu \right)$ 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 ${\nu }}$ 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 $\lambda$ and its SI unit is meter (m).

First, the energy required to produce one hydrogen atom in joule can be calculated as follows:

$E\left({\text{J}}\right)=\frac{{E\left( {{\text{kJ}}}\right)\times\left({\frac{{{{10}^3}{\text{ J}}}}{{{\text{kJ}}}}}\right)}}{{{{\text{N}}_{\text{A}}}}}$ …… (1)

Here, ${{\text{N}}_{\text{A}}}$ is an Avogadro's number and has a value $6.022 \times {10^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}$ .

Substitute $467{\text{ kJ/mol}}$ for $E\left( {{\text{kJ}}} \right)$ and $6.022 \times {10^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}$ for ${{\text{N}}_{\text{A}}}$ in equation (1).

$\begin{aligned}E\left( {\text{J}} \right)&=\frac{{\left({467{\text{ kJ/mol}}} \right) \times \left( {\frac{{{{10}^3}{\text{ J}}}}{{{\text{kJ}}}}}\right)}}{{6.022\times {{10}^{23}}{\text{ mo}}{{\text{l}}^{-1}}}}\\&=7.755\times{10^{-19}}{\text{ J}}\\\end{aligned}$

Thus the energy of a photon that can free one atom of hydrogen is $7.755 \times {10^{ - 19}}{\text{ J}}$ .

The expression of frequency and energy is as follows:

E=hv …… (2)

Here, is a frequency of photon and h is a Plank’s constant and has a value $\left({6.626\times{{10}^{-34}}{\text{ Js}}}\right)$ .

Rearrange equation (2) to calculate the frequency of the photon as follows:

$v=\frac{E}{h}$ …… (3)

Substitute $6.626\times{10^{-34}}{\text{ J}}\cdot{\text{s}}$ for h and $7.755 \times {10^{-19}}{\text{ J}}$ for E in equation (3).

$\begin{aligned}v&=\frac{{7.755\times{{10}^{-19}}{\text{ J}}}}{{6.626\times{{10}^{-34}}{\text{ Js}}}}\\&=1.17\times{10^{15}}{\text{}}{{\text{s}}^{-1}}\\\end{aligned}$

Thus the frequency of photon is $1.17\times{10^{15}}{\text{}}{{\text{s}}^{-1}}$ .

The expression to calculate the wavelength from energy of the photon is as follows:

$E = \frac{{h{\text{c}}}}{{\lambda }}}$ …… (4)

Here ${\lambda }}$ is a wavelength of a photon and c is a speed of light.

Rearrange equation (4) to calculate wavelength of the photon as follows:

${\lambda }}=\frac{{h{\text{c}}}}{E}$ …… (5)

Substitute $6.626 \times {10^{ - 34}}{\text{ J}} \cdot {\text{s}}$ for h, $3.0 \times {10^8}{\text{ m/s}}$ for c and $7.755 \times {10^{ - 19}}{\text{ J}}$ for E in equation (5).

$\begin{aligned}{\lambda}}=\frac{{\left({6.626\times {{10}^{-34}}{\text{ J}}\cdot {\text{s}}} \right)\left( {3.0 \times {{10}^8}{\text{ m/s}}}\right)}}{{\left({7.755\times {{10}^{ - 19}}{\text{ J}}} \right)}}\\=2.563\times {10^{-7}}{\text{m}}\\\end{aligned}$