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QJEEM19Ph3750: JEE Main 2019 Physics - Class 12 JEEM19Ph3750 Solution: In a Frank-Hertz experiment, an electron of energy 5.6 eV…

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 :

Multiple Choice Options

  • A 2020 nm
  • B 250 nm
  • C 1700 nm
  • D 220 nm

Step-by-Step Verified Solution

The minimum wavelength of emitted photons is

λ\lambda = 12405.6−0.7nm{{1240} \over {5.6 - 0.7}}nm = 250 nm

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