An electron in the hydrogen atom makes a transition,an energy state of principal quantom numbers ni to n2 state if the photon emitted has a wavelength 434nm what is the value of ni

To determine the value of ni, we can use the formula for calculating the wavelength of the emitted photon during a transition in the hydrogen atom:

1/λ = R(1/nf^2 - 1/ni^2)

Where:
- λ is the wavelength of the photon emitted (434 nm),
- R is the Rydberg constant (approximately 1.097 × 10^7 m^-1),
- nf is the final principal quantum number (n2 in this case),
- ni is the initial principal quantum number that we want to find.

Rearranging the equation, we can solve for ni:

1/ni^2 = 1/λ * (1/nf^2) + 1/ni^2

1/ni^2 = 1/λ * (1/nf^2)

ni^2 = λ * nf^2

ni = √(λ * nf^2)

Now, plug in the values and calculate ni:

ni = √(434 nm * (n2^2))

Finally, compute the result by substituting the value of n2:

ni = √(434 nm * (2^2))
= √(434 nm * 4)
= √(1736 nm)
≈ 41.67

Therefore, the approximate value of ni for the given transition is 41.67.