Question

In an X-ray photoelectron experiment, a photon of wavelength 121 pm ejects an electron from the inner shell of an atom and it emerges with a speed of 56.9 Mm s−1 . Calculate the binding energy of the electron.

Answer 1

Answer:

binding energy will be $0.1633\times 10^{-15}J$

Explanation:

We have given wavelength of photon $\lambda =121pm=121\times 10^{-12}m$

Velocity of light $c=3\times 10^8m/sec$

Plank's constant $h=6.6\times 10^{-34}Js$

So energy of photon $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{121\times 10^{-12}}=1.636\times 10^{-15}J$

Mass of electron $m=9.1\times 10^{-31}kg$

Velocity of electron is given $v=56.9\times 10^6m/sec$

So kinetic energy of electron $KE=\frac{1}{2}mv^2=\frac{1}{2}\times 9.1\times 10^{-31}\times (56.9\times 10^6)^2=1.473\times 10^{-15}J$

So binding energy = plank's energy - kinetic energy

$=1.636\times 10^{-15}-1.473\times 10^{-15}=0.1633\times 10^{-15}J$

So binding energy will be $0.1633\times 10^{-15}J$