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electric fields ions

https://isaacphysics.org/questions/identifying_ions?stage=a_level

can anyone give me tips for part e?

ive tried dividing the C/m ratio by the mass of the protons and neutrons and dividing it by 2, not really sure how to figure it out.
The ion is a "singly-charged positive atomic ion". From this you can deduce the charge.

You have the charge to mass ratio. From this, you have q2Np×mp=CM \frac{q}{2N_{p} \times m_{p}} = CM , where q is the charge of the ion, Np is the number of protons, mp is the mass of a proton and CM is the charge-mass ratio you calculated in the previous part. Rearrange this to get Np.

//Q: why do you have 2Np×mp 2N_{p}\times m_{p} on the denominator?
(edited 1 year ago)
Reply 2
Original post by MouldyVinegar
The ion is a "singly-charged positive atomic ion". From this you can deduce the charge.

You have the charge to mass ratio. From this, you have q2Np×mp=CM \frac{q}{2N_{p} \times m_{p}} = CM , where q is the charge of the ion, Np is the number of protons, mp is the mass of a proton and CM is the charge-mass ratio you calculated in the previous part. Rearrange this to get Np.

//Q: why do you have 2Np×mp 2N_{p}\times m_{p} on the denominator?


why is it 2Np×mp 2N_{p}\times m_{p} on the denominator? if there are equal numbers of protons and neutrons would it not just be 2*N_p since Np=mp


ended up getting it ty
(edited 1 year ago)
We have the charge to *mass* ratio. 2*Np does not give the mass, that only gives you the number of particles. You need to multiply by the mass of each particle, (which we take to be the same in the question) hence the denominator is 2*Np*mp .

In hindsight I probably should have written qMass=CM \frac{q}{Mass} = CM , and then you just need to figure out the Mass in terms of Np.

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