Question

An electron with energy 13.2 eV collides with an element, and the electron emerges with energy 4.5 eV. What wavelength of light would also be emitted along with the electron?

Answer 1

Answer: $140\ nm$

Explanation:

Given

Initial kinetic energy is $K_i=13.2\ eV$

Final kinetic energy is $K_f=4.5\ eV$

Change in kinetic energy is $\Delta K=K_i-K_f$

$\Rightarrow \Delta K=13.2-4.5\\\Rightarrow \Delta K=8.7\ eV$

Energy is given by

$E=\dfrac{hc}{\lambda}$

Equate it to change in kinetic energy

$\Rightarrow 8.7=\dfrac{hc}{\lambda}\\\\\Rightarrow \lambda =\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{8.7\times 1.6\times 10^{-19}}\\\\\Rightarrow \lambda =1.4\times 10^{-7}\ m\\\Rightarrow \lambda=140\ nm$

Therefore, the wavelength associated with it is $140\ nm$.