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

The problem states that for a hydrogen atom, the total energy of an electron in the first excited state is \(-3.4 \, \text{eV}\). We need to find the kinetic energy (K.E.) of this electron and express it in the form of \(x \times 10^{-1} \, \text{eV}\).

Concept Used:

According to the Bohr model for the hydrogen atom, the total energy (E) of an electron in a specific orbit is the sum of its kinetic energy (K.E.) and potential energy (P.E.). The relationships between these quantities are given by:

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

From these, a direct relationship between the total energy and the kinetic energy can be derived:

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

This means that the kinetic energy of the electron is the negative of its total energy. Since kinetic energy must be a positive value, the total energy of a bound electron is always negative.

Step-by-Step Solution:

Step 1: Identify the given total energy of the electron.

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

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

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

The formula is:

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

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

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

Final Computation & Result:

Step 4: Express the calculated kinetic energy in the required format.

The problem asks for the value of \(x\) where the kinetic energy is \(x \times 10^{-1} \, \text{eV}\).

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

Solving for \(x\):

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

The value of \(x\) is an integer, so no rounding is necessary.

The value of \(x\) is 34.

Answer 2

The energy of an electron in the first excited state for a hydrogen atom is given by:

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

For hydrogen, the kinetic energy (K.E.) in an orbit is given by:

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

Thus,

\[ x = -\left(-\frac{3.4}{2}\right) \times 10 = 34 \, \text{eV} \]