calculate the energy emitted when electron of 1 g-atom of hydrogen undergo transition giving the spectral lines of lowest energy in the visible region of its atomic spectra

To calculate the energy emitted when an electron in a hydrogen atom transitions between energy levels, we can use the Rydberg formula for hydrogen. This formula allows us to determine the wavelengths of the spectral lines produced when an electron moves from a higher energy level to a lower one. The lowest energy transitions that produce visible light in hydrogen occur when the electron falls to the second energy level (n=2) from higher levels (n=3, 4, etc.).

Understanding the Rydberg Formula

The Rydberg formula is expressed as:

1/λ = R_H (1/n1² - 1/n2²)

Where:

• λ is the wavelength of the emitted light.
• R_H is the Rydberg constant, approximately 1.097 x 10^7 m^-1.
• n1 is the lower energy level (for visible light, n1 = 2).
• n2 is the higher energy level (n2 = 3, 4, etc.).

Calculating the Wavelengths

For the transitions that result in visible light, we consider:

• Transition from n=3 to n=2 (Balmer series, H-alpha line)
• Transition from n=4 to n=2 (H-beta line)

Let's calculate the wavelength for the transition from n=3 to n=2:

1/λ = R_H (1/2² - 1/3²)

Substituting the values:

1/λ = 1.097 x 10^7 m^-1 (1/4 - 1/9)

Calculating the fractions:

1/λ = 1.097 x 10^7 m^-1 (9/36 - 4/36) = 1.097 x 10^7 m^-1 (5/36)

Now, simplifying:

λ = 36/(5 x 1.097 x 10^7) m

Calculating λ gives us approximately 656 nm, which corresponds to the red light in the visible spectrum.

Calculating the Energy Emitted

To find the energy emitted during this transition, we can use the formula:

E = hc/λ

Where:

• E is the energy in joules.
• h is Planck's constant (6.626 x 10^-34 J·s).
• c is the speed of light (3.00 x 10^8 m/s).

Substituting the values:

E = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / (656 x 10^-9 m)

Calculating this gives:

E ≈ 3.03 x 10^-19 J

Energy for 1 g-atom of Hydrogen

Now, since we are considering 1 g-atom of hydrogen, we need to find out how many hydrogen atoms are in 1 gram. The molar mass of hydrogen is approximately 1 g/mol, which means there are about 6.022 x 10^23 atoms in 1 gram.

To find the total energy emitted when all these atoms undergo the same transition, we multiply the energy per transition by the number of atoms:

Total Energy = E x Number of Atoms

Substituting the values:

Total Energy = (3.03 x 10^-19 J)(6.022 x 10^23)

Calculating this gives:

Total Energy ≈ 182 J

Thus, the energy emitted when 1 g-atom of hydrogen undergoes the transition to produce the lowest energy spectral lines in the visible region is approximately 182 joules. This illustrates how atomic transitions can release significant amounts of energy, especially when considering large quantities of atoms.