QJEEM23Phf88d: JEE Main 2023 Physics - Class 12 JEEM23Phf88d Solution: Experimentally it is found that 12.8   eV energy is…

The binding energy of an electron in a hydrogen atom is given by the formula:

$$E=\frac{k{e}^2}{2r}$$

where:

• $E$ is the energy of the electron,
• $k$ is Coulomb's constant ( $9 \times 10^9 \, \text{Nm}^2/\text{C}^2$ ),
• $e$ is the charge of the electron ( $1.6 \times 10^{-19} \, \text{C}$ ), and
• $r$ is the radius of the orbit.

In this scenario, the energy $E$ required to separate a hydrogen atom into a proton and an electron is given as $12.8 \, \text{eV}$ , which needs to be converted into joules using the conversion factor $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$ . So,

$$12.8\text{eV}=12.8\times 1.6\times {10}^{-19}\text{J}$$

We can then substitute the given values into the energy equation and solve for $r$ :

$$12.8\times 1.6\times {10}^{-19}\text{J}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{2r}$$

Solving for $r$ , we get:

$$r=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{2\times 12.8\times 1.6\times {10}^{-19}}$$

This simplifies to:

$$r=\frac{9\times {10}^{-10}}{16}$$

Comparing this with the given form of the radius, which is $\frac{9}{x} \times 10^{-10}$ , we find that the value of $x$ is 16.