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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.

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.

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

$E=\frac{hc}{位}$

Where, $位$is the wavelength, h is the Plank鈥檚 constant, c is the speed of light that is $3脳{10}^8\mathrm{m}/\mathrm{s}$, 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

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

Hence, the required wavelength is 253 nm.