According to Bohr's theory, the electronic energy of an electron in the `n^(th)` orbit is given by `E_(n) = (-2.17 xx 10^(-18))xx(z^2)/(n^(2)) J`
Calculate the longest wavelength of light that will be needed in remove an electron from the third Bohr orbit of `He^(o+)`

Explanation

To solve the problem of calculating the longest wavelength of light needed to remove an electron from the third Bohr orbit of He⁺, we can follow these steps:

### Step 1: Determine the energy of the electron in the third orbit (E₃)
According to Bohr's theory, the energy of an electron in the nth orbit is given by the formula:

\[ E_n = -\frac{2.17 \times 10^{-18} \, \text{J} \cdot z^2}{n^2} \]

For Helium ion (He⁺), the atomic number \( z = 2 \) and for the third orbit \( n = 3 \).

Substituting these values into the formula:

\[ E_3 = -\frac{2.17 \times 10^{-18} \cdot (2^2)}{3^2} \]

Calculating this gives:

\[ E_3 = -\frac{2.17 \times 10^{-18} \cdot 4}{9} = -\frac{8.68 \times 10^{-18}}{9} = -9.6444 \times 10^{-19} \, \text{J} \]

### Step 2: Calculate the energy required to remove the electron (ΔE)
The energy required to remove the electron from the third orbit to infinity (where the energy is 0) is given by:

\[ \Delta E = E_{\infty} - E_3 \]

Since \( E_{\infty} = 0 \):

\[ \Delta E = 0 - (-9.6444 \times 10^{-19}) = 9.6444 \times 10^{-19} \, \text{J} \]

### Step 3: Use the energy to find the wavelength (λ)
The relationship between energy and wavelength is given by the equation:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3.00 \times 10^8 \, \text{m/s} \) (speed of light)

Rearranging the equation to solve for wavelength \( \lambda \):

\[ \lambda = \frac{hc}{\Delta E} \]

Substituting the values:

\[ \lambda = \frac{(6.626 \times 10^{-34}) \cdot (3.00 \times 10^8)}{9.6444 \times 10^{-19}} \]

Calculating this gives:

\[ \lambda = \frac{1.9878 \times 10^{-25}}{9.6444 \times 10^{-19}} \approx 2.061 \times 10^{-7} \, \text{m} \]

### Final Answer
Thus, the longest wavelength of light needed to remove an electron from the third Bohr orbit of He⁺ is approximately:

\[ \lambda \approx 2.061 \times 10^{-7} \, \text{m} \, \text{or} \, 206.1 \, \text{nm} \]

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