In hydrogen atom, an electron jumps from 3d orbit to the 2nd orbit. Calculate the wavelength of the radiatoin emitted. `(h = 6.63 xx 10^(-34) J sec)`

Explanation

To solve the problem of calculating the wavelength of the radiation emitted when an electron in a hydrogen atom jumps from the 3d orbit (n=3) to the 2nd orbit (n=2), we can follow these steps:

### Step 1: Calculate the energy difference (ΔE) between the two orbits.
The energy of an electron in a hydrogen atom can be calculated using the formula:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

For n=2 and n=3, we can calculate the energies:

\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

Now, we find the energy difference:

\[ \Delta E = E_3 - E_2 = (-1.51 \, \text{eV}) - (-3.4 \, \text{eV}) = 1.89 \, \text{eV} \]

### Step 2: Convert the energy difference from eV to Joules.
To convert eV to Joules, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):

\[ \Delta E = 1.89 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.024 \times 10^{-19} \, \text{J} \]

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

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

Where:
- \(h\) is Planck's constant \(= 6.63 \times 10^{-34} \, \text{J s}\)
- \(c\) is the speed of light \(= 3 \times 10^8 \, \text{m/s}\)

Rearranging for λ gives:

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

### Step 4: Substitute the values into the formula.
Now we can substitute the values into the equation:

\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{3.024 \times 10^{-19} \, \text{J}} \]

Calculating this gives:

\[ \lambda = \frac{1.989 \times 10^{-25} \, \text{J m}}{3.024 \times 10^{-19} \, \text{J}} \approx 6.58 \times 10^{-7} \, \text{m} = 6580 \, \text{Å} \]

### Final Answer:
The wavelength of the radiation emitted when the electron jumps from the 3d orbit to the 2nd orbit is approximately **6580 Å**.

---