Two neutral particles are kept 1m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible. `h=6.63xx10^(-34)Js`.

Explanation

We know that initial charge on both is zero and hence we may assume zero electric potential energy initially.
`U_1=0`
When one electron is transferred , then both acquire equal and opposite charge having same magnitude as that of electron , so electric potential energy now can be written as follows :
`U_2=1/(4piepsilon_0)(q_1q_2)/r=9xx10^9xx((1.6xx10^(-19))(-1.6xx10^(-19)))/1`
`=-23.04xx10^(-29)J`
When one more electron is transferred , then charge on both becomes double and hence magnitude of electric potential energy will be 4 times to that of `U_2`.
`U_3=4xxU_2=-92.16xx10^(-19)J`
`U_1-U_2=hc//lamda_1`
`implies23.04xx10^(-29)=(6.63xx10^(-34)xx3xx10^8)/(23.0.4xx10^(-29))`
`=0.863xx10^3m`
`lamda_1=863m`
We can similarly proceed for the next wavelength as follows :
`U_2-U_3=hc//lamda_1`
`implies69.12xx10^(-29)=(6.63xx10^(-34)xx3xx10^8)/lamda_2`
`implieslamda_2=(6.63xx10^(-34)xx3xx10^8)/(69.12xx10^(29))`
`=0.288xx10^3m`
`implies lamda_1=288m`