Question

.An electron in the hydrogen atom makes a transition from a principal energy state ni to an energy state n=4. If the photon emitted has a wavelength of 734 nm, calculate the initial energy state ni? Constants: h= 6.63x10-34 J.S, C= 3x10 8 m/s and RH = 2.18x10-18 J.

Answer 1

The initial energy state ni can be calculated using the given equation for the energy of a hydrogen atom. By determining the energy of the emitted photon, we can substitute it into the equation to find ni. The calculation yields an approximation of ni to be 1.62.

Step-by-step explanation:

To calculate the initial energy state, we can use the equation for the energy of a hydrogen atom:

E = -RH*(1/ni^2)

Solving for ni:

ni = sqrt(-(E/RH))

Given the wavelength of the emitted photon, we can determine its energy using the equation:

E = (hc) / λ

Substituting the given values:

E = (6.63x10^-34 J.s * 3x10^8 m/s) / (734x10^-9 m)

Calculating the energy:

E = 2.71x10^-18 J

Finally, substituting this value of E into the equation for ni:

ni = sqrt(-(2.71x10^-18 J / 2.18x10^-18 J))

Calculating ni:

ni = 1.62

Therefore, the initial energy state ni is approximately 1.62.