The frequency of radiation emiited when the electron falls n =4 to n=1 in a hydrogen atom will be ( given ionization energy of `H= 2.18 xx 10 ^(-18)J "atom "^(-1) and h= 6.625 xx 10 ^(-34)Js)`

Options

A. `1.54xx 10^(15)s^(-1)`
B. `1.03 xx 10^(15)s^(-1)`
C. `3.08xx 10^(15 )s^(-1)`
D. `2.00xx10^(15) s^(-1)`

Answer

To find the frequency of radiation emitted when an electron falls from n = 4 to n = 1 in a hydrogen atom, we can follow these steps:

### Step 1: Calculate the Energy Levels
The energy of an electron in a hydrogen atom at a given principal quantum number n is given by the formula:

\[ E_n = -\frac{E}{n^2} \]

where \(E\) is the ionization energy of hydrogen, given as \(2.18 \times 10^{-18} \, \text{J}\).

### Step 2: Calculate \(E_1\) and \(E_4\)
1. For \(n = 1\):
\[ E_1 = -\frac{2.18 \times 10^{-18}}{1^2} = -2.18 \times 10^{-18} \, \text{J} \]

2. For \(n = 4\):
\[ E_4 = -\frac{2.18 \times 10^{-18}}{4^2} = -\frac{2.18 \times 10^{-18}}{16} = -1.3625 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the Change in Energy (\(\Delta E\))
The change in energy when the electron transitions from \(n = 4\) to \(n = 1\) is given by:

\[ \Delta E = E_1 - E_4 \]

Substituting the values:

\[ \Delta E = -2.18 \times 10^{-18} - (-1.3625 \times 10^{-19}) = -2.18 \times 10^{-18} + 1.3625 \times 10^{-19} \]

Calculating this gives:

\[ \Delta E = -2.04375 \times 10^{-18} \, \text{J} \]

### Step 4: Calculate the Frequency (\( \nu \))
Using Planck's equation, which relates energy and frequency:

\[ E = h \nu \]

Rearranging gives:

\[ \nu = \frac{\Delta E}{h} \]

Substituting the values (\(h = 6.626 \times 10^{-34} \, \text{Js}\)):

\[ \nu = \frac{2.04375 \times 10^{-18}}{6.626 \times 10^{-34}} \]

Calculating this gives:

\[ \nu \approx 3.08 \times 10^{15} \, \text{Hz} \]

### Final Answer
The frequency of radiation emitted when the electron falls from \(n = 4\) to \(n = 1\) in a hydrogen atom is approximately:

\[ \nu \approx 3.08 \times 10^{15} \, \text{Hz} \]