Question

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

• A. $1700\text{nm}$
• B. $220\text{nm}$
• C. $250\text{nm}$
• D. $2020\text{nm}$

Answer 1

The correct option is C $250\text{nm}$

The energy supplied by the electron to the mercury atoms is given by,

$\Delta E=5.5-0.7=4.9\text{eV}$

Therefore, the minimum wavelength of photons emitted by mercury atoms,

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

$\Rightarrow \lambda \simeq 250\text{nm}$

Hence, option $\left(D\right)$ is correct.