Question

For hydrogen atom, energy of an electron in first excited state is $-3.4 \, \text{eV}$, K.E. of the same electron of hydrogen atom is $x \, \text{eV}$.Value of $x$ is ____ $\times 10^{-1} \, \text{eV}$. (Nearest integer)

Answer 1

Correct Answer: 34

For a hydrogen atom's electron in the first excited state, the total energy is given as \(-3.4 \, \text{eV}\). The objective is to determine the electron's kinetic energy (K.E.) and present it in the format \(x \times 10^{-1} \, \text{eV}\).

Concept Used:

The Bohr model for hydrogen atoms posits that the total energy (E) of an electron in an orbit is the sum of its kinetic energy (K.E.) and potential energy (P.E.). The following relationships are established:

\[ \text{K.E.} = -\frac{1}{2} \text{P.E.} \] \[ E = \text{K.E.} + \text{P.E.} \]

From these equations, a direct correlation between total energy and kinetic energy can be deduced:

\[ \text{K.E.} = -E \]

This indicates that the kinetic energy is the inverse of the total energy. Since kinetic energy must be positive, the total energy of a bound electron is inherently negative.

Step-by-Step Solution:

Step 1: Note the provided total energy of the electron.

The total energy of the electron in the first excited state is:

\[ E = -3.4 \, \text{eV} \]

Step 2: Apply the formula linking kinetic energy and total energy.

The applicable formula is:

\[ \text{K.E.} = -E \]

Step 3: Substitute the value of E into the formula to compute the kinetic energy.

\[ \text{K.E.} = -(-3.4 \, \text{eV}) \] \[ \text{K.E.} = 3.4 \, \text{eV} \]

Final Computation & Result:

Step 4: Express the computed kinetic energy in the specified format.

The problem requires the value of \(x\) such that the kinetic energy equals \(x \times 10^{-1} \, \text{eV}\).

\[ x \times 10^{-1} \, \text{eV} = 3.4 \, \text{eV} \]

Solve for \(x\):

\[ x = \frac{3.4}{10^{-1}} = 3.4 \times 10 \] \[ x = 34 \]

Since \(x\) is an integer, no approximation is needed.

The value of \(x\) is 34.