In hy drogen spectrum, a hydrogen
atom emits a photon of wavelength
1027 `Å` its angular momentum
changes by

Options

A. `(h)/(pi)`
B. `(h)/(2 pi)`
C. `(3 h)/(2 pi)`
D. `(2 h)/(pi)`

Explanation

To solve the problem of how much the angular momentum of a hydrogen atom changes when it emits a photon of wavelength 1027 Å, we can follow these steps:

### Step 1: Calculate the energy of the emitted photon
The energy of the emitted photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

First, convert the wavelength from angstroms to meters:
\[ 1027 \, \text{Å} = 1027 \times 10^{-10} \, \text{m} \]

Now, substitute the values into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{1027 \times 10^{-10} \, \text{m}} \]

Calculating this gives:
\[ E \approx 1.24 \times 10^{-6} \, \text{J} \]

To convert this energy into electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E \approx \frac{1.24 \times 10^{-6}}{1.6 \times 10^{-19}} \approx 12.1 \, \text{eV} \]

### Step 2: Determine the initial and final energy states
For a hydrogen atom, the energy levels are given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

Assuming the electron transitions from the third energy level (\( n = 3 \)) to the ground state (\( n = 1 \)):
- Initial energy (\( E_3 \)):
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \, \text{eV} \]

- Final energy (\( E_1 \)):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]

### Step 3: Calculate the change in energy
The change in energy (\( \Delta E \)) when the electron transitions from \( n = 3 \) to \( n = 1 \) is:
\[ \Delta E = E_1 - E_3 = (-13.6) - (-1.51) = -12.1 \, \text{eV} \]

### Step 4: Calculate the angular momentum
The angular momentum for an electron in a given orbit is given by:
\[ L_n = \frac{nh}{2\pi} \]

- For \( n = 3 \):
\[ L_3 = \frac{3h}{2\pi} \]

- For \( n = 1 \):
\[ L_1 = \frac{1h}{2\pi} \]

### Step 5: Calculate the change in angular momentum
The change in angular momentum (\( \Delta L \)) is:
\[ \Delta L = L_3 - L_1 = \frac{3h}{2\pi} - \frac{1h}{2\pi} = \frac{(3 - 1)h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi} \]

### Final Answer
The change in angular momentum when the hydrogen atom emits a photon of wavelength 1027 Å is:
\[ \Delta L = \frac{h}{\pi} \]

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