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.

Answer 1

Explanation:

Given data in the question :

Gap between the two neutral particles, r = 1 m

Potential electric energy

$E_{1}=\frac{\mathrm{kq}^{2}}{\mathrm{r}}=\mathrm{kq}^{2}$

where $k=\frac{1}{4 \pi \varepsilon_{0}}$

Photon Energy  

$E 2=\frac{\mathrm{hc}}{\lambda}$

where λ = light wavelength  

         h is  Planck's constant

         c = light speed  

Here, E1 = E2

$k q^{2}=\frac{h c}{\lambda}$

$\lambda=\frac{h c}{k q^{2}}$

To be optimum, charging λ should be minimum for wavelength.

$q=e=1.6 \times 10^{-19} \mathrm{C}$

Wavelength of maximum,  

$\lambda=\frac{h c}{k q^{2}}$  

  $=\frac{6.63 \times 3 \times 10^{-34} \times 10^{8}}{9 \times 10^{2} \times 1.6^{2} \times 10^{-38}}$

  $=0.863 \times 103=863 \mathrm{m}$

Lesser wave frequency,

$\lambda=\frac{6.63 \times 3 \times 10^{-34} \times 10^{8}}{9 \times 4 \times 10^{4} \times 1.6^{2} \times 10^{-38}}$

  = $\frac{863}{4}$

  =215.74 m