What electron transition in a hydrogen atom, ending in the orbit 5, will produce light of wavelength 3040 ?
3040 what? angstroms. Whatever it is convert to meters. You say ENDING in orbit 5 so I assume that means it starts further out.
(1/wavelength) = R*(1/25 - 1/x^2)
R = 1.09737E7
To determine the electron transition in a hydrogen atom that produces light of a specific wavelength, you need to use the formula for calculating the wavelength of light emitted during a transition in the hydrogen atom's energy levels. This formula is known as the Rydberg formula:
1/λ = R_H (1/n1^2 - 1/n2^2),
where λ is the wavelength of light emitted, R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m^-1), and n1 and n2 are the initial and final energy levels of the electron, respectively.
In this case, the wavelength of the light emitted is given as 3040 nm (or 3.04 × 10^-6 meters). Plugging this value into the equation:
1/3.04 × 10^-6 = 1.097 × 10^7 (1/n1^2 - 1/5^2).
Simplifying the equation:
1/n1^2 - 1/25 = 3.28 × 10^13.
To solve for n1, you need to find the common denominator:
25/n1^2 - 1/25 = 3.28 × 10^13.
Multiply everything by 25n1^2:
25 - n1^2 = 82 × 10^14n1^2.
Rearranging the equation:
(82 × 10^14 + 1)n1^2 = 25.
Divide both sides by (82 × 10^14 + 1):
n1^2 = 25 / (82 × 10^14 + 1).
Taking the square root of both sides:
n1 = √(25 / (82 × 10^14 + 1)).
Calculating n1 using the equation:
n1 ≈ 8.377.
Since n1 must be a positive integer, approximate it to the nearest whole number:
n1 ≈ 8.
Therefore, the electron transition in a hydrogen atom that ends in the orbit labeled as 5 will produce light of wavelength 3040 nm.