Question

A Incident X-rays in a Compton scattering experiment have a wavelength of 0.042 nm and scattered X rays have a wavelength of 0.044 nm what is the kinetic energy of recoiled electrons? B A hydrogen atom is ionized with a 400-nm photon. If the ejected electron previously was in the state with n - 3, find its kinetic energy after it has been ejected.

Answer 1

ai, Ina.dank You, waive angth λ.-0.04Qnm scattereol-Y24 4aeleno? th λ교-o.o44 nm ehci'; aden hc_hc 「33.8-22.チ + 19.86 x 10 -เชี่

ai, Ina.dank You, waive angth λ.-0.04Qnm scattereol-Y24 4aeleno? th λ교-o.o44 nm ehci'; aden hc_hc 「33.8-22.チ + 19.86 x 10 -เชี่

Answer 2

SOLUTION :

K.E. of recoiled electrons

= K.E. of incident electron - K. E. of scattered electron

= h c / wavelength of incident ray - h c / wavelength of scattered ray

= h c / (0.042 nm) - h c / (0.044 nm)

= h c * 0.002/(0.042*0.044) * 10^9

= (6.62 * 10^(-34) * 3 * 10^8) * 1.082251 * 10^9

= 2.149 * 10^(-16) J ( ANSWER).