If the wavelength of photon emitted from an electron jump `n=4` to `n=2` in a `H-` like species is `1216 Å` then the species is `:`

Options

A. `H-` atom
B. `He^(+)` ion
C. `Li^(2+)`ion
D. `Be^(3+)`ion

Explanation

To determine the species of the hydrogen-like ion based on the given photon emission wavelength, we will follow these steps:

### Step 1: Calculate the Energy of the Photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
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 (\(3.00 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

First, convert the wavelength from angstroms to meters:
\[ \lambda = 1216 \, \text{Å} = 1216 \times 10^{-10} \, \text{m} \]

Now, substituting the values into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{1216 \times 10^{-10} \, \text{m}} \]

Calculating this gives:
\[ E \approx 1.64 \times 10^{-18} \, \text{J} \]

To convert this energy into electron volts (1 eV = \(1.602 \times 10^{-19} \, \text{J}\)):
\[ E \approx \frac{1.64 \times 10^{-18}}{1.602 \times 10^{-19}} \approx 10.24 \, \text{eV} \]

### Step 2: Identify the Transition
The energy calculated corresponds to the transition of an electron from a higher energy level to a lower energy level. In hydrogen-like species, the energy levels can be described by the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \(Z\) is the atomic number of the ion.

The energy difference between levels \(n=4\) and \(n=2\) can be calculated as:
\[ \Delta E = E_4 - E_2 = -\frac{Z^2 \cdot 13.6}{4^2} + \frac{Z^2 \cdot 13.6}{2^2} \]
\[ = Z^2 \cdot 13.6 \left(\frac{1}{4} - \frac{1}{16}\right) = Z^2 \cdot 13.6 \left(\frac{4 - 1}{16}\right) = Z^2 \cdot 13.6 \cdot \frac{3}{16} \]

Setting this equal to the energy we calculated:
\[ Z^2 \cdot 13.6 \cdot \frac{3}{16} = 10.24 \]

### Step 3: Solve for Z
Rearranging gives:
\[ Z^2 = \frac{10.24 \cdot 16}{13.6 \cdot 3} \]
Calculating the right-hand side:
\[ Z^2 = \frac{163.84}{40.8} \approx 4 \]
Thus, \(Z = 2\).

### Conclusion
Since \(Z = 2\), the species is Helium ion (He\(^+\)), which is a hydrogen-like species.

### Final Answer
The species is **Helium ion (He\(^+\))**.

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