InterviewSolution
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InterviewSolution
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Find the velocity of the excited hydrogen atoms if the first line of the Lyman series is displaced by Delta lambda= 0.20 nm when their radiation is observed at an angle theta=45^(@) to their motion direction. |
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Answer» Solution :Since `DELTA lambda(= 0.20nm)lt lt lambda(= 121nm)` of the first LYMAN line of `H` atom, we need not worry about `V^(2)//c^(2)` effects. Then `omega'=(omega)/(1-betacos theta),beta=(v)/(c )` Hence `1- beta COS theta=(omega)/(omega')=(lambda')/(lambda)` or `beta cos theta=1-(lambda')/(lambda)=(Delta lambda)/(lambda)` But `omega=(3)/(4)R so lambda=(2pic)/(omega)=(8pic)/(3R)` Hence `v=cbeta=(3RDeltalambda)/(8pi cos theta)` Substitution gives `( cos theta=(1)/(sqrt(2)))` `v= 7.0xx10^(5)m//s` |
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