Be sure to answer all parts. 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. Frequency = 7.05 × 10 -32 s−1 Wavelength = 4.26 × 10 -25 m Minimum energy = × 10 kJ/photon

Respuesta :

Answer:

[tex]7.7549\times 10{19}\ J[/tex]

[tex]1.17037\times 10^{15}\ Hz[/tex]

[tex]2.56329\times 10^{-7}\ m[/tex]

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]

[tex]\nu[/tex] = Frequency

[tex]\lambda[/tex] = Wavelength

The minimum energy is given by

[tex]E=\frac{1\ mol}{N_A}\\\Rightarrow E=467\times 10^{3}\times \frac{1}{6.022\times 10^{23}}\\\Rightarrow E=7.7549\times 10{19}\ J[/tex]

The minimum energy is [tex]7.7549\times 10{19}\ J[/tex]

The energy of a photon is given by

[tex]E=h\nu\\\Rightarrow \nu=\frac{E}{h}\\\Rightarrow \nu=\frac{467\times 10^{3}\frac{1}{6.022\times 10^{23}}}{6.626\times 10^{-34}}\\\Rightarrow \nu=1.17037\times 10^{15}\ Hz[/tex]

The frequency of the photon is [tex]1.17037\times 10^{15}\ Hz[/tex]

Wavelength is given by

[tex]\lambda=\frac{c}{\nu}\\\Rightarrow \lambda=\frac{3\times 10^8}{1.17037\times 10^{15}}\\\Rightarrow \lambda=2.56329\times 10^{-7}\ m[/tex]

The wavelength is [tex]2.56329\times 10^{-7}\ m[/tex]