In a hypothetical atom, mass of electron is doubled, value
of atomic number is `Z = 4`. Find wavelength of photon when this electron jumps
from 3rd excited state to 2nd orbit.

Explanation

To solve the problem of finding the wavelength of a photon emitted when an electron jumps from the third excited state to the second orbit in a hypothetical atom where the mass of the electron is doubled and the atomic number \( Z = 4 \), we can follow these steps:

### Step 1: Understand the Energy Levels
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \times Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 2: Calculate the Energy for the 3rd Excited State (n = 4)
For the third excited state, where \( n = 4 \):
\[ E_4 = -\frac{13.6 \times (4)^2}{(4)^2} = -\frac{13.6 \times 16}{16} = -13.6 \, \text{eV} \]
Since the mass of the electron is doubled, we need to multiply the energy by 2:
\[ E_4 = -13.6 \times 2 = -27.2 \, \text{eV} \]

### Step 3: Calculate the Energy for the 2nd Orbit (n = 2)
For the second orbit, where \( n = 2 \):
\[ E_2 = -\frac{13.6 \times (4)^2}{(2)^2} = -\frac{13.6 \times 16}{4} = -54.4 \, \text{eV} \]
Again, multiplying by 2 due to the doubled mass of the electron:
\[ E_2 = -54.4 \times 2 = -108.8 \, \text{eV} \]

### Step 4: Calculate the Energy Difference
The energy difference \( \Delta E \) when the electron jumps from \( n = 4 \) to \( n = 2 \) is:
\[ \Delta E = E_2 - E_4 = (-108.8) - (-27.2) = -108.8 + 27.2 = -81.6 \, \text{eV} \]

### Step 5: Calculate the Wavelength of the Photon
The wavelength \( \lambda \) of the emitted photon can be calculated using the formula:
\[ \lambda = \frac{12375 \, \text{Å eV}}{\Delta E} \]
Substituting the value of \( \Delta E \):
\[ \lambda = \frac{12375}{81.6} \approx 151.65 \, \text{Å} \]

### Final Answer
The wavelength of the photon emitted when the electron jumps from the third excited state to the second orbit is approximately:
\[ \lambda \approx 151.65 \, \text{Å} \]

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