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A photon with initial frequency 10^(11)Hz scatters off an electron at rest. Its final frequency is 0.9 xx 10^(11)Hz. Calculate the speed of the scattered electron (h = 6.63 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg). |
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/incident-1039786" style="font-weight:bold;" target="_blank" title="Click to know more about INCIDENT">INCIDENT</a> energy = Scattered energy + K.E. of the scattered electron <br/> `hv_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) = hv_(2) + (1)/(2) <a href="https://interviewquestions.tuteehub.com/tag/mv-1082193" style="font-weight:bold;" target="_blank" title="Click to know more about MV">MV</a>^(2)` <br/> or `(1)/(2) mv^(2) = h (v_(1) - v_(2))` <br/> or `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>^(2) = (2h(v_(1) - v_(2)))/(m) = (2 xx 6.63 xx 10^(-34) kg m^(2) s^(-1) (10^(11) - 0.9 xx 10^(11)) s^(-1))/(9.1 xx 10^(-31) kg)` <br/> `= (2 xx 6.63 xx 10^(-34) xx 0.1 xx 10^(11))/(9.1 xx 10^(-31)) m^(2) s^(-2) = 14571428 m^(2) s^(-2)` <br/> or `v = sqrt(14571428) ms^(-1) = 3.8 xx 10^(3) ms^(-1)`</body></html> | |