Question

Two neutral particles are kept 1 m 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.

(Use h = 6.63 × 10^-34J-s = 4.14 × 10^-15 eV-s, c = 3 × 10⁸ m/s and me = 9.1 × 10^-31kg)

Answer 1

Given:-

Distance between the two neutral particles, r = 1 m

Electric potential energy,

`E_1 = (kq^2)/r = kq^2`

where `K = 1/(4pi∈_0)`

Energy of photon,

`E_2 = (hc)/λ`,

where λ = wavelength of light
          h = Planck's constant
          c = speed of light

Here, E₁ = E₂

`therefore kq^2 = (hc)/λ`

`⇒ λ = (hc)/(kq^2)`

For wavelength, λ, to be maximum, charge q should be minimum.

`q = e = 1.6 xx 10^-19  "C"`

Maximum wavelength,

`λ = (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 10^3 = 863  "m"`

Next smaller wave length,

`λ = (6.63 xx 3 xx 10^-34 xx 10^8)/(9 xx 10^4 xx 4 xx (1.6)^2 xx 10^-38)`

`= 863/4`

`= 215.74  "m"`