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

Here ` r = 1m `
` Energy = kq^(2) / R = kq^(2) / 1 `
Now , kq^(2) / 1 = hc / lambda`
or , `lambda = hc/ kq^(2) `
For max `' lambda '` , 'q' should be minimum For minimum
'e '` = 1.6 xx 10^(-19) C`
Maximum `lambda = hc / kq^(2) `
` 6.63 xx 3 xx 10^(-34) xx 10^(8) / 9 xx 10^(2) xx (1.6)^(2) xx 10^(-38) `
`= 0.863 xx 10^(2)= 863 m`
For next smaller wave length
` 6.63 xx 3 xx 10^(-34) xx 10^(8) / 9 xx 10^(4) xx (1.6)^(4) xx 10^(-38) `
` = 863 / 4 `
`= 215.74m `.