Question

The space between two parallel metal plates is filled with an element in a gaseous stale. Electrons leave one plate at negligible speed and are accelerated toward the other by a potential difference$\mathbf{∆}\mathbf{V}$applied between the plates. As $\mathbf{∆}\mathbf{V}$is increased from 0, the electron current increases more or less linearly, but when $\mathbf{∆}\mathbf{V}$reaches 4.9 V , the current drops precipitously. From nearly 0 , it builds again roughly linearly as $\mathbf{∆}\mathbf{V}$is increased beyond 4.9 V .

(a) How can the presence of the gas explain these observations?

(b) The Gas emits pure “light” when $\mathbf{∆}\mathbf{V}$exceeds 4.9 V . What is its wavelength?

Short answer

(a) Energy is not enough for the quantum jump when $∆\mathrm{V}$is low, but at high $∆\mathrm{V}$, the jump occurs and current increases.

(b) The Wavelength is .

Step-by-step solution

A concept:

A quantum jump refers to the transition of electrons from one state to another. When an electron gains energy enough for transition, it jumps to a higher state, and it loses energy while it jumps down from that state.

Electrons gain kinetic energy along the way despite being at rest when they start to move from one plate.

(a) Behaviour of Electron current

During the energy gain, if the energy is insufficient for their quantum jump, no behaviour is observed.

But in case they get energy sufficient to cause a jump, they cause the jump and dissipate a large amount of energy and they speed up again.

As applied voltage further increases, they acquire more energy for exiting and they speed up more hence the current gain increases.

(b) Wavelength of the light:

If we assume, the kinetic energy of the electron is the energy of the jump, the energy of the jump will be 4.9 eV .

For calculating the wavelength, we will use the equation given below

$E=\frac{hc}{\lambda }$

Where, $\lambda$is the wavelength, h is the Plank’s constant, c is the speed of light that is $3\times {10}^8\mathrm{m}/\mathrm{s}$, and E is the energy of photon.

Consider the known data as below.

$$\begin{gathered} \mathrm{hc}=1240\mathrm{eV}.\mathrm{nm} \\ \mathrm{E}=4.9\mathrm{eV} \end{gathered}$$

Substitute these values into equation (1), and you have

$$\begin{gathered} \frac{1240\mathrm{eV}.\mathrm{nm}}{\mathrm{\lambda }}=4.9\mathrm{eV} \\ \mathrm{\lambda }=253\mathrm{nm} \end{gathered}$$

Hence, the required wavelength is 253 nm.