# 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.

Navjyot Kalra
10 years ago
Sol. r = 1 m Energy = kq^2/R = kq^2/1 Now, kQ^2/1 = hc/λ or λ = hc/kq^2 For max ‘λ’, ‘q’ should be min, For λ hc/kq^2 – 0.863 * 10^3 = 863 m. For next smaller wavelength = 6.63 * 3 *10^-34 * 10^8/9 * 10^9 *(1.6 * 2)^2 * 10^-38 = 863/4 = 215.74 m
Dhawal Patil
24 Points
7 years ago
As charge is quantized, the least amount of charge that can be transferred is e.
According to the question, energy lost is completely transferred in the form of the photon.
Please note that the energy is lost as conservative forces are considered to be negative if they are attractive in nature.
As electric potential energy  =
$-\frac{1}{4\pi\epsilon_o}\!\!\frac{{q}^2}{r}$
And energy of the photon = $h\nu$

We have :
$-\frac{1}{4\pi\epsilon_o}\!\!\frac{{q}^2}{r} = h\nu$
Sustituting, $\nu = \frac{c}{\lambda}$ and discarding the minus sign, we get:
$\frac{1}{4\pi\epsilon_o}\!\!\frac{q^2}{r} = \frac{hc}{\lambda}$
which implies,
$\lambda=\frac{hc}{\frac{1}{4\pi\epsilon_o}\!\!\frac{q^2}{r}}$
Also, $\lambda_{n^{th}largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(n.1\!\cdot\!6.10^{-19})^2}$
$\therefore\lambda_{largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(1\!\cdot\!6.10^{-19})^2}$$\therefore\lambda_{largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(2.1\!\cdot\!6.10^{-19})^2}=\frac{\lambda_{largest}}{4}$