According to Bohr's theory , the electronic energy of hydrogen atom is the nth Bohr's orbit is given by
`E_(n) = (-21.76 xx 10^(-19))/(n^(2)) J`
Calculate the longest wavelength of electron from the third Bohr's of the `He^(+)` ion

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the energy of the electron in the third Bohr's orbit of the He⁺ ion.

According to Bohr's theory, the energy of an electron in the nth orbit of a hydrogen-like atom is given by the formula:

\[ E_n = \frac{-21.76 \times 10^{-19}}{n^2} \text{ J} \]

For the He⁺ ion, which has an atomic number \( Z = 2 \), we need to modify the formula to account for the atomic number:

\[ E_n = \frac{-21.76 \times 10^{-19} \times Z^2}{n^2} \text{ J} \]

Substituting \( Z = 2 \) and \( n = 3 \):

\[ E_3 = \frac{-21.76 \times 10^{-19} \times (2^2)}{3^2} \]

Calculating this:

\[ E_3 = \frac{-21.76 \times 10^{-19} \times 4}{9} = \frac{-87.04 \times 10^{-19}}{9} = -9.64 \times 10^{-19} \text{ J} \]

### Step 2: Use the energy to find the wavelength of the emitted photon.

The energy of a photon is related to its wavelength by the equation:

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

Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \text{ m/s} \)),
- \( \lambda \) is the wavelength.

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

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

Substituting the values:

\[ \lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{9.64 \times 10^{-19} \text{ J}} \]

Calculating the numerator:

\[ hc = 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \text{ J m} \]

Now substituting this into the equation for \( \lambda \):

\[ \lambda = \frac{1.9878 \times 10^{-25}}{9.64 \times 10^{-19}} \approx 2.06 \times 10^{-7} \text{ m} \]

### Final Answer:

The longest wavelength of the electron from the third Bohr's orbit of the He⁺ ion is:

\[ \lambda \approx 2.06 \times 10^{-7} \text{ m} \]

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