Energy of an electron is given by `E = - 2.178 xx 10^-18 J ((Z^2)/(n^2))`. Wavelength of light required to excite an electron in a hydrogen atom from level `n = 1` to `n = 2` will be
`(h = 6.62 xx 10^-34 Js` and `c = 3.0 xx 10^8 ms^-1`).

Options

A. `1.214 xx 10^-7 m`
B. `2.816 xx 10^-7 m`
C. `6.500 xx 10^-7 m`
D. `8.500 xx 10^-7 m`

Explanation

To find the wavelength of light required to excite an electron in a hydrogen atom from level \( n = 1 \) to \( n = 2 \), we can follow these steps:

### Step 1: Calculate the energy levels
The energy of an electron in a hydrogen atom is given by the formula:
\[ E = -2.178 \times 10^{-18} \, \text{J} \left( \frac{Z^2}{n^2} \right) \]
For hydrogen, \( Z = 1 \).

#### Energy at \( n = 1 \):
\[ E_1 = -2.178 \times 10^{-18} \, \text{J} \left( \frac{1^2}{1^2} \right) = -2.178 \times 10^{-18} \, \text{J} \]

#### Energy at \( n = 2 \):
\[ E_2 = -2.178 \times 10^{-18} \, \text{J} \left( \frac{1^2}{2^2} \right) = -2.178 \times 10^{-18} \, \text{J} \left( \frac{1}{4} \right) = -5.445 \times 10^{-19} \, \text{J} \]

### Step 2: Calculate the energy difference (\( \Delta E \))
The energy difference between the two levels is:
\[ \Delta E = E_2 - E_1 \]
Substituting the values we calculated:
\[ \Delta E = -5.445 \times 10^{-19} \, \text{J} - (-2.178 \times 10^{-18} \, \text{J}) = -5.445 \times 10^{-19} \, \text{J} + 2.178 \times 10^{-18} \, \text{J} \]
Calculating this gives:
\[ \Delta E = 1.6335 \times 10^{-18} \, \text{J} \]

### Step 3: Relate energy difference to wavelength
The energy of a photon is also given by:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3.0 \times 10^8 \, \text{ms}^{-1} \) (speed of light)

Setting the two expressions for energy equal gives:
\[ \Delta E = \frac{hc}{\lambda} \]
Rearranging to find \( \lambda \):
\[ \lambda = \frac{hc}{\Delta E} \]

### Step 4: Substitute values and calculate \( \lambda \)
Substituting the known values:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.0 \times 10^8 \, \text{ms}^{-1})}{1.6335 \times 10^{-18} \, \text{J}} \]
Calculating the numerator:
\[ hc = 6.626 \times 10^{-34} \times 3.0 \times 10^8 = 1.9878 \times 10^{-25} \, \text{Jm} \]
Now substituting back into the equation for \( \lambda \):
\[ \lambda = \frac{1.9878 \times 10^{-25}}{1.6335 \times 10^{-18}} \approx 1.216 \times 10^{-7} \, \text{m} \]

### Final Result
The wavelength of light required to excite the electron from \( n = 1 \) to \( n = 2 \) is approximately:
\[ \lambda \approx 1.216 \times 10^{-7} \, \text{m} \text{ or } 121.6 \, \text{nm} \]

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