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  • Stuck here, help me understand: In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV . The minimum wavelength of photons emitted by mercury atoms is close to :-11226

 In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV . The minimum wavelength of photons emitted by mercury atoms is close to : 

  • Option 1)

    220 nm

  • Option 2)

    250 nm

  • Option 3)

    1700 nm

  • Option 4)

    2020 nm 

Answers (1)

best_answer

 

Frank Hertz Experiment -

Frank and Hertz had proposed that the 4.9 V characteristic of their experiments was due to ionization of mercury atoms by collisions with the flying electrons emitted at the cathode.

- wherein

E_{1} = 5.6 eV

E_{2} = 0.7 eV

\Delta E = 5.6 -0.7 =4.9eV

\lambda =\frac{hc}{\Delta E} = \frac{1240}{4.9}mm

\lambda =250nm

 


Option 1)

220 nm

Option 2)

250 nm

Option 3)

1700 nm

Option 4)

2020 nm 

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