Answer:
Explanation:
Given that
Beam current (i)=23.3µA
And the time to strike(t)=28s
Also, a fundamental charge e=1.602×10^-19C
Then, the charge quantity is given as,
q=it
Then, q=23.3×10^-6×28
q=6.524×10^-4C
Also, the number of electron N is given as
q=Ne
Therefore, N=q/e
So, N=6.524×10-4/1.602×10^-19
N=4.072×10^15
There are 4.072×10^15 electrons strike the tube screen every 28 s.
When The fundamental charge is 1.602 × 10−19 C So, There are [tex]4.072\times 10\wedge 15[/tex] electrons that strike the tube screen every 28 s.
Given that information as per the question
Beam current (i) is =23.3µA
And also the time to strike(t)=28s
Also, a fundamental charge e=[tex]1.602\times10\wedge -19C[/tex]
Then, the charge quantity is given as,
q is =it
Then, q=[tex]23.3\times 10\wedge-6×28[/tex]
q is =[tex]6.524\times10\wedge -4C[/tex]
Also, When the number of electron N is given as
Now, q=Ne
Therefore, N is =q/e
So, N is = [tex]6.524\times10-4/1.602\times10\wedge-19[/tex]
N is = [tex]4.072\times 10\wedge 15[/tex]
Therefore, There are [tex]4.072\times 10\wedge 15[/tex] electrons that strike the tube screen every 28 s.
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