Question

In a Frank-Hertz experiment, an electron of energy 5.6 V passes through mercury
vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted
by meaçury atoms is close to
(a)
1700 nm (b)220nm(c) 250 nm
(4) 2020 nm​

Answer 1

Answer:

Wavelength of the photon emitted is 250 nm

Explanation:

Energy level $E_{1}=5.6 eV$

and $E_{2}=0.7 eV$

Energy retained by the mercury vapour is

$\Delta E= E_{1}-E_{2}\\\Delta E=5.6-0.7=4.9 eV$

Wavelength of the photon emitted

take the value of hc in eV is =1240 nm

$\lambda=\frac{hc}{\Delta E} \\\\\lambda=\frac{1240}{4.9} \\\\\lambda=250 nm$

Wavelength of the photon emitted is 250 nm