ionization energy

[205743684]

6. Calculation the ionization energy of a rubidium atom, given that radiation of wavelength
58.4 nm produces electrons with a speed of 2450 km/s

does anyone know how to solve this

[Anne Mariano 3F]

First, you need to find the energy of the photon by using E=hv. Then the energy of an electron is found by using E=(1/2)mv^2. Ionization energy is then solved for by subtracting the kinetic energy from the photon's initial energy.

[Reza Hemmati 3L]

Hi!

I'm fairly sure that this is a photoelectric effect problem, where you have the formula E_photon = work function + KE_electron.

First, you can convert the wavelength 58.4nm into a value for the energy of the photon by using the equation E = hc/l(where l is lambda)
E = (6.626 x 10^-34)(3.0 x 10^8)/(58.4 x 10^-9 m) = 3.404 x 10^-18 J

Then, you can use the speed of the electron to find out its kinetic energy for the equation using KE = (1/2) mv^2 (for this one be careful, m is in kg and v in m/s)
KE = (1/2)(9.109 x 10^-31 kg)(2.450 x 10^6 m/s)^2 = 2.734 x 10^-18 J

Next, use these values for the original equation!
3.404 x 10^-18 J = work function + 2.734 x 10^-18 J
work function = 0.670 x 10^-18 J

In metals, the ionization energy(for gaseous state) is the same as the work function, so there is your answer!

Hope this helps! ^^

[Raizel Ferrer 1H]

I would first convert the given units to the appropriate units so they match. Convert the wavelength nm to m and the velocity from km/s to m/s. I would then use the equation E = hc/lambda to find the energy of the photon. Then, use an additional equation KE = 1/2mv^2 to calculate the energy of the electron. Use the values from both of these equations and plug it into E(photon) + work function = KE. The work function should then equal the ionization energy.