Answer with Explanation:
We are given that
q=+2 e
K.E=24 MeV=[tex]24\times 10^6e V[/tex]
[tex]1 MeV=10^6e V[/tex]
Current,I=260 nA=[tex]260\times 10^{-9} A[/tex]
a.Time,[tex]\Delta t[/tex]=3 s
We know that
Number of alpha particles,N=[tex]\frac{T\Delta t}{q}[/tex]
Using the formula
[tex]N=\frac{260\times 10^{-9}\times 3}{2\times 1.6\times 10^{-91}}[/tex]
Where [tex] e=1.6\times 10^{-19} C[/tex]
[tex]N=2.4375\times10^{12}[/tex]
b.Potential difference,V=[tex]\frac{K.E}{q}[/tex]
Using the formula
[tex]V=\frac{24\times 10^6 e}{2e}=12\times 10^6 V[/tex]
