Question

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

Answer 1

`7.7549* 10{19}\ J`

`1.17037* 10^(15)\ Hz`

`2.56329* 10^(-7)\ m`

Step-by-step explanation:

c = Speed of light =
`3* 10^8\ m/s`

`N_A` = Avogadro's number =
`6.022* 10^(23)`

`\\u` = Frequency

`\lambda` = Wavelength

The minimum energy is given by

`E=(1\ mol)/(N_A)\\\Rightarrow E=467* 10^(3)* (1)/(6.022* 10^(23))\\\Rightarrow E=7.7549* 10{19}\ J`

The minimum energy is
`7.7549* 10{19}\ J`

The energy of a photon is given by

`E=h\\u\\\Rightarrow \\u=(E)/(h)\\\Rightarrow \\u=(467* 10^(3)(1)/(6.022* 10^(23)))/(6.626* 10^(-34))\\\Rightarrow \\u=1.17037* 10^(15)\ Hz`

The frequency of the photon is
`1.17037* 10^(15)\ Hz`

Wavelength is given by

`\lambda=(c)/(\\u)\\\Rightarrow \lambda=(3* 10^8)/(1.17037* 10^(15))\\\Rightarrow \lambda=2.56329* 10^(-7)\ m`

The wavelength is
`2.56329* 10^(-7)\ m`