Energy of an electron is given by `E= -2.178 xx 10^(-18)J ((z^(2))/(n^(2)))`. The 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, C= 3.0 xx 10^(8) ms^(-1)`)

Options

A. `1.214 xx 10^(-7)m`
B. `2.816 xx 10^(-7)m`
C. `6.560 xx 10^(-7)m`
D. `8.500 xx 10^(-7)m`

Answer

To solve the problem of finding 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 at n=1 and n=2
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, the atomic number \( z = 1 \).

1. **Calculate \( E_1 \) (for \( n = 1 \))**:
\[ E_1 = -2.178 \times 10^{-18} \, \text{J} \left( \frac{1^2}{1^2} \right) = -2.178 \times 10^{-18} \, \text{J} \]

2. **Calculate \( E_2 \) (for \( n = 2 \))**:
\[ E_2 = -2.178 \times 10^{-18} \, \text{J} \left( \frac{1^2}{2^2} \right) = -2.178 \times 10^{-18} \, \text{J} \left( \frac{1}{4} \right) = -0.5445 \times 10^{-18} \, \text{J} \]

### Step 2: Calculate the Energy Difference
The energy difference \( \Delta E \) between the two levels is given by:
\[ \Delta E = E_2 - E_1 \]
Substituting the values:
\[ \Delta E = (-0.5445 \times 10^{-18}) - (-2.178 \times 10^{-18}) = 1.6335 \times 10^{-18} \, \text{J} \]

### Step 3: Use the Energy-Wavelength Relation
The relation between energy and wavelength is 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)

Rearranging for \( \lambda \):
\[ \lambda = \frac{hc}{\Delta E} \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.0 \times 10^{8} \, \text{ms}^{-1})}{1.6335 \times 10^{-18} \, \text{J}} \]

Calculating the numerator:
\[ hc = 6.626 \times 10^{-34} \times 3.0 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{Jm} \]

Now substituting into the wavelength equation:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{1.6335 \times 10^{-18}} \approx 1.216 \times 10^{-7} \, \text{m} = 121.6 \, \text{nm} \]

### Final Answer
The wavelength of light required to excite the electron from level n=1 to n=2 in a hydrogen atom is approximately **121.6 nm**.

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