In a Frank-Hertz experiment, an electron of energy 5.6 eV
Question:

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

  1. (1) $1700 \mathrm{~nm}$

  2. (2) $2020 \mathrm{~nm}$

  3. (3) $220 \mathrm{~nm}$

  4. (4) $250 \mathrm{~nm}$


Correct Option: , 4

Solution:

(4) Using, wavelength, $\lambda=\frac{12375}{\Delta \mathrm{E}}$

or, $\lambda=\frac{12375}{4.9}=250 \mathrm{~nm}$

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