Among the first lines of Lyman,Balmer, paschen and Brackett series in hydrogen atomic spectra,which has the highest energy?

Options

A. Lyman
B. Balmer
C. Paschen
D. Bracket

Answer

To determine which of the first lines of the Lyman, Balmer, Paschen, and Brackett series in hydrogen atomic spectra has the highest energy, we can use the formula for the energy of a photon emitted during an electron transition in a hydrogen atom:

\[ E = R_H \cdot h \cdot c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R_H \) is the Rydberg constant,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

### Step 1: Calculate Energy for Lyman Series
For the Lyman series, the transition is from \( n_2 = 2 \) to \( n_1 = 1 \):
\[ E_{Lyman} = R_H \cdot h \cdot c \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ E_{Lyman} = R_H \cdot h \cdot c \left( 1 - \frac{1}{4} \right) \]
\[ E_{Lyman} = R_H \cdot h \cdot c \left( \frac{3}{4} \right) \]

### Step 2: Calculate Energy for Balmer Series
For the Balmer series, the transition is from \( n_2 = 3 \) to \( n_1 = 2 \):
\[ E_{Balmer} = R_H \cdot h \cdot c \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ E_{Balmer} = R_H \cdot h \cdot c \left( \frac{1}{4} - \frac{1}{9} \right) \]
\[ E_{Balmer} = R_H \cdot h \cdot c \left( \frac{9 - 4}{36} \right) \]
\[ E_{Balmer} = R_H \cdot h \cdot c \left( \frac{5}{36} \right) \]

### Step 3: Calculate Energy for Paschen Series
For the Paschen series, the transition is from \( n_2 = 4 \) to \( n_1 = 3 \):
\[ E_{Paschen} = R_H \cdot h \cdot c \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
\[ E_{Paschen} = R_H \cdot h \cdot c \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ E_{Paschen} = R_H \cdot h \cdot c \left( \frac{16 - 9}{144} \right) \]
\[ E_{Paschen} = R_H \cdot h \cdot c \left( \frac{7}{144} \right) \]

### Step 4: Calculate Energy for Brackett Series
For the Brackett series, the transition is from \( n_2 = 5 \) to \( n_1 = 4 \):
\[ E_{Brackett} = R_H \cdot h \cdot c \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \]
\[ E_{Brackett} = R_H \cdot h \cdot c \left( \frac{1}{16} - \frac{1}{25} \right) \]
\[ E_{Brackett} = R_H \cdot h \cdot c \left( \frac{25 - 16}{400} \right) \]
\[ E_{Brackett} = R_H \cdot h \cdot c \left( \frac{9}{400} \right) \]

### Step 5: Compare Energies
Now we have the energies for all series:
- \( E_{Lyman} = R_H \cdot h \cdot c \cdot \frac{3}{4} \)
- \( E_{Balmer} = R_H \cdot h \cdot c \cdot \frac{5}{36} \)
- \( E_{Paschen} = R_H \cdot h \cdot c \cdot \frac{7}{144} \)
- \( E_{Brackett} = R_H \cdot h \cdot c \cdot \frac{9}{400} \)

To determine which is the highest, we can compare the fractions:
- \( \frac{3}{4} = 0.75 \)
- \( \frac{5}{36} \approx 0.1389 \)
- \( \frac{7}{144} \approx 0.0486 \)
- \( \frac{9}{400} = 0.0225 \)

Clearly, \( E_{Lyman} \) has the highest energy.

### Conclusion
The first line of the Lyman series has the highest energy among the given series.