An electron in a Bohr orbit of hydrogen atom with quantum number n has an angular momentum `4.2176xx10^(-34)kg-m^(2)//sec`.If the electron drops from this level to the next lower level , the wavelength of this lines is

Options

A. `5.86 Å`
B. `3.26Å`
C. `4.66Å`
D. `4.77Å`

Answer

To solve the problem of finding the wavelength of the photon emitted when an electron in a hydrogen atom drops from a higher energy level to a lower one, we can follow these steps:

### Step 1: Determine the Quantum Number (n)
The angular momentum \( L \) of an electron in a Bohr orbit is given by the formula:
\[ L = n \cdot \frac{h}{2\pi} \]
where \( h \) is the Planck constant. We are given:
\[ L = 4.2176 \times 10^{-34} \, \text{kg m}^2/\text{s} \]
Using \( h = 6.626 \times 10^{-34} \, \text{Js} \), we can rearrange the formula to find \( n \):
\[ n = \frac{L \cdot 2\pi}{h} \]
Substituting the values:
\[ n = \frac{4.2176 \times 10^{-34} \cdot 2 \cdot 3.14}{6.626 \times 10^{-34}} \]
Calculating this gives:
\[ n \approx 4 \]

### Step 2: Identify the Transition
The electron drops from \( n = 4 \) to \( n = 3 \).

### Step 3: Calculate the Energy Difference
The energy levels of the hydrogen atom are given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Calculating the energies for \( n = 4 \) and \( n = 3 \):
\[ E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV} \]
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \, \text{eV} \]
The energy difference \( \Delta E \) when the electron transitions from \( n = 4 \) to \( n = 3 \) is:
\[ \Delta E = E_3 - E_4 = -1.51 - (-0.85) = -1.51 + 0.85 = -0.66 \, \text{eV} \]

### Step 4: Calculate the Wavelength
Using the formula relating energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{E} \]
Substituting \( h = 6.626 \times 10^{-34} \, \text{Js} \), \( c = 3 \times 10^8 \, \text{m/s} \), and converting \( E \) to joules (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E = 0.66 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.056 \times 10^{-19} \, \text{J} \]
Now substituting into the wavelength formula:
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.056 \times 10^{-19}} \]
Calculating this gives:
\[ \lambda \approx 1.88 \times 10^{-6} \, \text{m} = 1880 \, \text{nm} \]

### Final Answer
The wavelength of the emitted photon when the electron drops from \( n = 4 \) to \( n = 3 \) is approximately \( 1880 \, \text{nm} \).

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