Calculate the energy of the hydrogen atom in the states n = 4 and n = 2. Determine the frequency and wavelength of the emitted radiation in a transition from n = 4 to n=2 state. Is this radiation visible?
Correct Answer - `1xx10^(16) Hz`
For frequency `2.5xx10^(16) Hz, h v=hv_(o)+KE_(max)`
`:. h(2.5xx10^(16))=hv_(o)+KE_(max) ...(i)`
For frequency `4xx10^(16)` Hz,
`h(4xx10^(16))=hv_(o)+2KE_(max) ...(2)`
Multiply `eq^(n)` (1) by `2` and substract `eq^(n)` (2) from it....
Correct Answer - `v=7.3xx10^(14) Hz`, visible spectrum
`DeltaE_(6 to 2)=hv rArr v=(3.022xx1.6xx10^(-19))/(6.625xx10^(-34))=7.3xx10^(14) Hz`
This frequency lies in visible spectrum.