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

Question

In a Frank - Hertz experiment, an electron of energy 5.6 &#8201; e V passes through mercury vapour and emerges with an energy 0.7 &#8201; e V . The minimum wavelength of photons emitted by mercury atoms is close to:

Aicharya · Accepted Answer

When electron pass through the mercury vapor, it losses some of its energy. The loss in

K E

of electron

= (5 \cdot 6 - 0.7) e V

= 4.9 e V

∴

energy of radiation emitted

= 4.9 e V

∴

wavelength of radiation,

λ = \frac{1.24 \times 10^{4}}{4.9} A \approx 250 n m

In a Frank - Hertz experiment, an electron of energy 5.6   e V passes through mercury vapour and…

Solution

In a Frank - Hertz experiment, an electron of energy 5.6 &#8201; e V passes through mercury vapour and…

Solution

In a Frank - Hertz experiment, an electron of energy 5.6 e V passes through mercury vapour and…