I have no idea how to do this problem, and I've been searching through my book for an hour. Help please?
An energy of 2.0x10^2 kJ/mol is required to cause a cesium atom on a metal surface to lose an electron. Calculate the longest possible wavelength of light that can ionize a cesium atom. In what region of the electromagnetic spectrum is this radiation found?
How many kilocalories are released by freezing 4.6 moles of H20?
You are not likely to obtain an answer by piggy backed questions; I suggest you post a new question at the top of the board.
E = hc/wavelength
Substitute h, c, and E and solve for wavelength. E is the energy/atom; to obtain that use 200,000 J/mole and convert to an atom by dividing by 6.022E23.
Is that visible light? Convert to Angstroms. 1 A = 1E-8 cm and 4000-7000 A is visible.
Sure, I'll be happy to help you with this problem!
To calculate the longest possible wavelength of light that can ionize a cesium atom, we need to use the equation for energy of a photon:
E = hc/λ
Where:
E is the energy of the photon,
h is Planck's constant (6.626 x 10^-34 J·s),
c is the speed of light (3.00 x 10^8 m/s), and
λ is the wavelength of the light.
Given that the energy required to ionize a cesium atom is 2.0 x 10^2 kJ/mol, we need to convert it to J by multiplying by 1000 and dividing by Avogadro's number (6.022 x 10^23):
Energy (E) = (2.0 x 10^2 kJ/mol) * (1000 J/kJ) / (6.022 x 10^23 mol^-1)
Now we can solve for the wavelength (λ):
λ = hc / E
Substituting the values of h, c, and E into the equation:
λ = (6.626 x 10^-34 J·s) * (3.00 x 10^8 m/s) / [(2.0 x 10^2 kJ/mol) * (1000 J/kJ) / (6.022 x 10^23 mol^-1)]
Calculating this expression will give us the longest possible wavelength of light that can ionize a cesium atom.
To determine in what region of the electromagnetic spectrum this radiation is found, we can refer to the standard classification:
- Radio waves: Wavelengths longer than about 10^3 m (1 km)
- Microwaves: Wavelengths from about 10^2 m (1 m) to 10^-3 m (1 mm)
- Infrared: Wavelengths from about 10^-3 m (1 mm) to 7 x 10^-7 m (700 nm)
- Visible light: Wavelengths from about 7 x 10^-7 m (700 nm) to 4 x 10^-7 m (400 nm)
- Ultraviolet: Wavelengths from about 4 x 10^-7 m (400 nm) to 10^-9 m (1 nm)
- X-rays: Wavelengths from about 10^-9 m (1 nm) to 10^-11 m (0.1 nm)
- Gamma rays: Wavelengths shorter than about 10^-11 m (0.1 nm)
Now we can determine the region of the electromagnetic spectrum the calculated wavelength falls into.
I hope this explanation helps you solve the problem!