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Quantum phenomena

http://filestore.aqa.org.uk/subjects/AQA-PHYA1-QP-JUN14.PDF

help pls
question 3c
(edited 8 years ago)
Reply 1
Avatar for Kyx
Kyx
6
lambda = h/mv
Reply 2
Avatar for thefatone
thefatone
OP
6
Original post by Kyx
lambda = h/mv


so i have v=hmvv=\dfrac{h}{mv}
then
v=6.63×10−34(9.11×10−31×207)×2.9×10−9v=\dfrac {6.63 \times 10^{-34}}{\left( 9.11 \times 10^{-31} \times 207 \right) \times 2.9 \times 10^{-9}}
v=1200ms−1 v=1200ms^{-1}

thanks
(edited 8 years ago)
Reply 3
Avatar for Kyx
Kyx
6
Original post by thefatone
so i have v=hmvv=\dfrac{h}{mv}
then
v=6.63×10−34(9.11×10−31×207)×2.9×10−9v=\dfrac {6.63 \times 10^{-34}}{\left( 9.11 \times 10^{-31} \times 207 \right) \times 2.9 \times 10^{-9}}
v=1200ms−1 v=1200ms^{-1}

thanks


np

although it's lambda equals, not v equals
Original post by thefatone
so i have v=hmvv=\dfrac{h}{mv}
then
v=6.63×10−34(9.11×10−31×207)×2.9×10−9v=\dfrac {6.63 \times 10^{-34}}{\left( 9.11 \times 10^{-31} \times 207 \right) \times 2.9 \times 10^{-9}}
v=1200ms−1 v=1200ms^{-1}

thanks


Or can say velocity is propotional to 1/m, so if m is 207 times as big can multiply velocity in question by 1/207. Same thing just less chance of making an algebraic error, and not dependent on any rounding you might have done earlier
Reply 5
Avatar for thefatone
thefatone
OP
6
Original post by Kyx
np

although it's lambda equals, not v equals


oops that's supposed to be rearranged

v=hmλv=\dfrac{h}{m\lambda}
Reply 6
Avatar for Kyx
Kyx
6
Original post by thefatone
oops that's supposed to be rearranged

v=hmλv=\dfrac{h}{m\lambda}


I thought so :smile:

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