Find the wavelength in a hydrogen spectrum between the range `500nm to 700nm`

Options

A. `540nm`
B. `580nm`
C. `654nm`
D. `696nm`

Answer

C

Explanation

The energy of photon of wavelength `500nm` is
` (hc)/lambda=(1242eV-nm)/(500nm)`
The energy of a proton of wavelength `700nm` is `(hc)=(1242eV-nm)/(700nm)=1.77eV`
The energy different Between the state involved in the transition should , `"therefore"` , be beteeen `1.77` ,
`eV` and `2.44 eV` . `n=3`to `n=2` fall in this
range ` lambda=(hc)/(DeltaE) =(1242eV-nm)/(1.9eV)=654nm`