The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is 3.4 eV. Determine the de- broglie wavelength associated with it.
Mass of electron=`9.1xx10^(-31)kg`,
Planck's constant=`6.63xx10^(-34)Js`.

Explanation

To determine the de Broglie wavelength associated with the electron orbiting in the first excited state of a hydrogen atom with a kinetic energy of 3.4 eV, we can follow these steps:

### Step 1: Convert Kinetic Energy from eV to Joules
The kinetic energy (KE) is given in electron volts (eV). We need to convert this to joules (J) using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).

\[ KE = 3.4 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 5.44 \times 10^{-19} \text{ J} \]
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