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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)
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Solution
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"`
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