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

Answer:

The voltage would be  $V_{F} = 8.41 V$      

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

                   $V = \frac{E}{e} = \frac{h c}{e \lambda }$

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

            $V_{F} = 4 * 2.107$

            $V_{F} = 8.41 V$