What is the wavelength of a photon emitted during a transition form the `n_(i) = 5` state to the `n_(f) = 2` state in the `H` atom?

Options

A. `568 nm`
B. `786 nm`
C. `434 nm`
D. `678 nm`

Answer

C

Explanation

Step `1. DeltaE = Z^(2)E_(1) ((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
`=(1)^(2) = (-2.18 xx 10^(-18)J) ((1)/(2^(2))-(1)/(5^(2)))`
`= - 4.58 xx 10^(-19)J atm^(-1)`
The negative sign indicates that enegry is released.
Step `2`.
Note `DeltaE = hv = (hc)/(lambda)`
`lambda = (ch)/(DeltaE) = ((300xx10^(8)ms^(-1))(6.63xx10^(-34)Js))/((4.58xx10^(-19)J))`
`4.34 xx 10^(-7)m`
`(4.34 xx 10^(-7)m) ((10^(9)nm)/(1m)) = 434 nm`