Modern Physics> Two neutral particles are kept 1 m apart....

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.

Simran Bhatia , 11 Years ago

Grade 11

2 Answers

Navjyot Kalra

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

Last Activity: 11 Years ago

Dhawal Patil

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, $\lambda_{n^{th}largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(n.1\!\cdot\!6.10^{-19})^2}$

$\therefore\lambda_{largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(1\!\cdot\!6.10^{-19})^2}$ $\therefore\lambda_{largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(2.1\!\cdot\!6.10^{-19})^2}=\frac{\lambda_{largest}}{4}$