Question

The first excited state of sodium decays to the ground state by emitting a photon of wavelength 590 nm. If sodium vapor is used for the Franck-Hertz experiment, at what voltage will the fourth current drop be recorded

Answer 1

The voltage would be
`V_(F) = 8.41 V`

Step-by-step explanation:

From the question we are told that

The wavelength is
`\lambda = 590 nm = 590*10^(-9) m`

Generally when a electron changes state to a higher state an energy(E) is given off and this is equivalent to eV

So the voltage of the electron can be evaluated as

`V = (E)/(e) = (h c)/(e \lambda )`

Where h is the planks constant with a constant value
`h = 6.625*10^(-34)`

c is the speed of light
`c = 3*10^8 m/s`

e is the value charge in one electron
`e = 1.602 *10^(- 19) C`

Substituting value

`V = \frac{(6.625 *10{-34} ( 3*10^(8)))}{(1.60*10^(-19) ) (590 *10^(-9))}`

`V= 2.107 V`

So the fourth current drop would be recorded when electron has change state four times and the voltage at this point can be mathematically evaluated as

`V_(F) = 4 * 2.107`

`V_(F) = 8.41 V`