In the emission spectrum of hydrogen, calculate the wavelength (in nm) of emitted light in the Lyman series corresponding to an initial quantum state of ni = 4

To calculate the wavelength of emitted light in the Lyman series corresponding to an initial quantum state of ni = 4 in the emission spectrum of hydrogen, we can use the Rydberg formula. The Rydberg formula is given by:

1/λ = R_H * (1/ni^2 - 1/nf^2)

Where λ is the wavelength of the emitted light, R_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m^-1), ni is the initial quantum state, and nf is the final quantum state.

In the Lyman series, the final quantum state nf is always equal to 1. Therefore, we can substitute nf = 1 into the Rydberg formula:

1/λ = R_H * (1/4^2 - 1/1^2)

Simplifying the equation:

1/λ = R_H * (1/16 - 1)

1/λ = R_H * (1/16 - 16/16)

1/λ = R_H * (-15/16)

Now, we can solve for λ by taking the reciprocal of both sides of the equation:

λ = 16/(-15 * R_H)

Substituting the value of R_H:

λ = 16/(-15 * 1.097 x 10^7 m^-1)

Calculating the value:

λ ≈ -9.524 x 10^-8 m

To convert the wavelength to nanometers, we can multiply by 10^9:

λ ≈ -9.524 x 10^-8 m * 10^9 nm/m

λ ≈ -95.24 nm

Therefore, the wavelength of the emitted light in the Lyman series corresponding to an initial quantum state of ni = 4 is approximately 95.24 nm.