In the emission spectrum of hydrogen what is the
wavelength of the light emitted by the transition
fro m n = 4 to n =2.
1/wavelength = R(1/2^2 - 1/4^2)
R = 1.09737E7
a hydrogen atom absorbs a photon of visible light n=2 to n=4 energy level. calculate a) the change in energy of the atom
b) the wave length (in nm) of the photon
3322
To find the wavelength of the light emitted by the transition from n = 4 to n = 2 in the emission spectrum of hydrogen, you can use the Rydberg formula:
1/λ = R * (1/n1^2 - 1/n2^2)
Where:
- λ is the wavelength of the emitted light
- R is the Rydberg constant (approximately 1.097 × 10^7 m^-1)
- n1 and n2 are the initial and final energy levels, respectively
In this case, n1 = 4 and n2 = 2. Plugging these values into the formula:
1/λ = R * (1/4^2 - 1/2^2)
Simplifying the equation:
1/λ = R * (1/16 - 1/4)
1/λ = R * (1/16 - 4/16)
1/λ = R * (-3/16)
Now we can calculate the value of 1/λ:
1/λ = -3R/16
To find the value of λ, we need to take the reciprocal of 1/λ:
λ = -16/(3R)
Substituting the value of the Rydberg constant:
λ = -16/(3 * 1.097 × 10^7 m^-1)
Calculating the value:
λ ≈ 6.56 × 10^-8 meters
Therefore, the approximate wavelength of the light emitted by the transition from n = 4 to n = 2 is 6.56 × 10^-8 meters.