Consider a hydrogen atom with its electron in the `n^(th)` orbital An electomagnetic radiation of wavelength `90 nm` is used to ionize the atom . If the kinetic energy of the ejected electron is `10.4 eV` , then the value of `n` is ( hc = 1242 eVnm)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the energy of the incoming photon
The energy of the photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant multiplied by the speed of light, given as \( 1242 \, \text{eV nm} \),
- \( \lambda \) is the wavelength of the photon, given as \( 90 \, \text{nm} \).

Substituting the values:
\[ E = \frac{1242 \, \text{eV nm}}{90 \, \text{nm}} = \frac{1242}{90} \approx 13.8 \, \text{eV} \]

### Step 2: Relate the energy of the photon to the ionization energy and kinetic energy
The total energy of the incoming photon is used to overcome the ionization energy of the electron in the hydrogen atom and provide it with kinetic energy. The relationship can be expressed as:
\[ E = \text{Ionization Energy} + \text{Kinetic Energy} \]
The ionization energy for a hydrogen atom in the \( n^{th} \) orbital is given by:
\[ \text{Ionization Energy} = -\frac{13.6 \, \text{eV}}{n^2} \]
Thus, we can write:
\[ 13.8 \, \text{eV} = -\frac{13.6 \, \text{eV}}{n^2} + 10.4 \, \text{eV} \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ 13.8 \, \text{eV} - 10.4 \, \text{eV} = -\frac{13.6 \, \text{eV}}{n^2} \]
Calculating the left side:
\[ 3.4 \, \text{eV} = -\frac{13.6 \, \text{eV}}{n^2} \]

### Step 4: Solve for \( n^2 \)
Multiplying both sides by \( -n^2 \) gives:
\[ -3.4 n^2 = 13.6 \]
Dividing both sides by -3.4:
\[ n^2 = \frac{13.6}{3.4} = 4 \]

### Step 5: Calculate \( n \)
Taking the square root of both sides:
\[ n = \sqrt{4} = 2 \]

### Final Answer
The value of \( n \) is \( 2 \).

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