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)
Correct Answer: 34
Approach Solution - 1
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.
Approach Solution -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,
\[ x = -\left(-\frac{3.4}{2}\right) \times 10 = 34 \, \text{eV} \]
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