Hydrogen gas in the atomic state is excited to an energy level such that the electrostatic polential energy becomes `-3.02 eV`
Now, the photoelectric plate hoving W=4.6 eV is exposed to the emission spectra of this gas. Assuming all the transition of be possible, find the minimum de-Broglie wavelength of ejected photoelectrons.

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Determine the Total Energy of the Excited State
The electrostatic potential energy (U) of the hydrogen atom in the excited state is given as -3.02 eV. The total energy (E) of the electron in the hydrogen atom is related to the potential energy by the formula:
\[ E = \frac{U}{2} \]
So we calculate:
\[ E = \frac{-3.02 \, \text{eV}}{2} = -1.51 \, \text{eV} \]

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