Question

In a frank-hertz experiment, an electron of energy \[5.6eV\] passes through mercury vapour and emerges with energy \[0.7eV\]. The minimum wavelength of photons emitted by mercury atoms is close to:-
\[{\text{A}}{\text{. }}2020{\text{ }}nm\]
\[{\text{B}}{\text{. }}220{\text{ }}nm\]
\[{\text{C}}{\text{. }}250{\text{ }}nm\]
\[{\text{D}}{\text{. }}1700{\text{ }}nm\]

Answer 1

Hints:In frank- hertz experiments, in a tube filled with mercury vapour is electrons accelerated. From the distance between the equidistant minima of the electron current in a variable opposing electric field, the excitation energy of mercury is determined.

Complete step by step answer:
The Frank hertz had proposed that their experiments are \[4.9{\text{ }}V\] characteristic due to ionization of mercury atoms by collisions with the flying electrons and it is emitted at the cathode.
The given value of an electron energy is \[5.6eV\] passes through the mercury vapour and \[0.7eV\] energy of emerges, and to find the minimum wavelength of photons emitted by mercury atoms is close to,
Energy retained by mercury vapour
The electron energy passes through the mercury vapour is ${E_1} = 5.6eV$
And the energy of emerges is ${E_2} = 0.7eV$
$\Delta E = 5.6 - 0.7$
$ \Rightarrow 4.9eV$
The minimum wavelength of photons emitted by mercury atoms, $\lambda = \dfrac{{hc}}{{\Delta E}}$
So we can apply the values, and we get
\[ \Rightarrow \dfrac{{1240}}{{4.9}}mm\]
Divided from the above value, and the answer is,
$ \Rightarrow \lambda = 250nm$
So, the correct answer is option \[\left( c \right)\].

Additional information:
The first electrical measurement is the franck- Hertz experiment and it was to clearly show the quantum nature of atoms, and thus it is transformed out of the world.
In April \[24\], \[1914\] it was presented to the German physical society in a paper by James Franck and Gustav hertz.
Frank and Hertz explained their experiment, In terms of elastic and inelastic collisions between the electrons and the mercury atoms. With the mercury atoms slowly moving electrons collide elastically. This means that the direction in which the electron is moving is altered by the collision, but its speed is unchanged.

Note:The franck-Hertz experiment confirms the Bohr model of that atom.
 When electrons in a potential field were found that were passed through mercury vapour they experienced an energy loss in distinct steps, and that the mercury gave an emission line at $\lambda = 254nm$.