A hydrogen spectrum is observed using a hydrogen lamp, an emission line with a wavelength of 398 nm is associated with excited electrons dropping down to the n = 2 shell. The initial n value for this transition is [Ans].

To find the initial quantum number (n), we can use the Rydberg formula for hydrogen:

1/λ = R * (1/n1² - 1/n2²)

where:

• λ is the wavelength of the light emitted,
• R is the Rydberg constant (approximately 1.097373 x 10^7 m^-1),
• n1 is the principal quantum number of the lower energy level,
• n2 is the principal quantum number of the higher energy level.

Given in the problem:

• λ = 398 nm = 398 x 10^-9 m (since 1 nm = 10^-9 m),
• n1 = 2 (the electron drops down to the n = 2 shell).

We can rearrange the Rydberg formula to solve for n2:

1/n2² = R/λ + 1/n1²

Substituting the given values:

1/n2² = (1.097373 x 10^7 m^-1) / (398 x 10^-9 m) + 1/2² 1/n2² = 2.76 x 10^15 + 0.25 1/n2² = 2.76 x 10^15 + 0.25 n2² = 1 / (2.76 x 10^15 + 0.25) n2 = sqrt(1 / (2.76 x 10^15 + 0.25))

Calculating the above expression will give the initial quantum number (n2).

The question is asking for the initial energy level (n value) of an electron in a hydrogen atom before it drops to the n = 2 shell, emitting light with a wavelength of 398 nm in the process.

To solve this, we can use the Rydberg formula for hydrogen:

1/λ = R * (1/n1² - 1/n2²)

where:

• λ is the wavelength of the emitted light,
• R is the Rydberg constant for hydrogen (approximately 1.097373 x 10^7 m^-1),
• n1 is the lower energy level (which is 2 in this case), and
• n2 is the higher energy level (which we're trying to find).

Rearranging the formula to solve for n2 gives us:

n2 = sqrt(1/(R * λ + 1/n1²))

Substituting the given values into this equation:

n2 = sqrt(1/(1.097373 x 10^7 m^-1 * 398 x 10^-9 m + 1/2²))

Solving this equation will give us the initial n value for this transition.