
The charge of an electron is \[1.6 \times {10^{ - 19}}\]C. How many electrons strike the screen of the cathode ray tube each second, when the beam current is 16mA?
A. \[{10^{ - 19}}\]
B. \[{10^{ - 17}}\]
C. \[{10^{18}}\]
D. \[{10^{17}}\]
Answer
161.1k+ views
Hint:The electric current is the rate of flow of charge per unit time. Total charge is the product of the number of particles and the charge carried by each particle.
Formula used:
\[i = \dfrac{q}{t}\]
where i is the magnitude of electric current when q charge flows across a cross-sectional area in time t.
\[q = ne\]
where q is the total charge contained by n number of electrons and e is the charge on an electron.
Complete step by step solution:
The charge on an electron is given as \[1.6 \times {10^{ - 19}}C\]
\[e = 1.6 \times {10^{ - 19}}C\]
The electric current is given as 16 mA. As we know that the electric current is the rate of flow of charge, so \[16 \times {10^{ - 3}}C\] charge flows across a cross-section per second.
Using the definition of electric current, we find total charge per second.
\[I = \dfrac{q}{t} \\ \]
\[\Rightarrow 16\,mA = \dfrac{q}{{1s}} \\ \]
\[\Rightarrow q = \left( {16 \times {{10}^{ - 3}}A} \right) \times \left( {1s} \right) \\ \]
\[\Rightarrow q = 16 \times {10^{ - 3}}C\]
So, the total charge striking the screen of the cathode ray tube is \[16 \times {10^{ - 3}}C\].
Let there be n electrons strike the screen of the cathode ray tube each second. Then the total charge striking the cathode ray tube per second will be,
\[q = ne\]
\[\Rightarrow q = 1.6 \times {10^{ - 19}}n\]
From the obtained charge,
\[1.6 \times {10^{ - 19}}n = 16 \times {10^{ - 3}} \\ \]
\[\Rightarrow n = \dfrac{{16 \times {{10}^{ - 3}}}}{{1.6 \times {{10}^{ - 19}}}} \\ \]
\[\Rightarrow n = {10^{17}}\]
Hence, \[{10^{17}}\] electrons strike the screen of the cathode ray per second when the current is 16 mA.
Therefore, the correct option is D.
Note: The electric current is a scalar quantity, so to find the rate of flow of charge we take the magnitude of the charge rather than considering whether charge is negative or positive. The charge on an electron is \[ - 1.6 \times {10^{ - 19}}C\] and the magnitude of the charge is \[1.6 \times {10^{ - 19}}C\].
Formula used:
\[i = \dfrac{q}{t}\]
where i is the magnitude of electric current when q charge flows across a cross-sectional area in time t.
\[q = ne\]
where q is the total charge contained by n number of electrons and e is the charge on an electron.
Complete step by step solution:
The charge on an electron is given as \[1.6 \times {10^{ - 19}}C\]
\[e = 1.6 \times {10^{ - 19}}C\]
The electric current is given as 16 mA. As we know that the electric current is the rate of flow of charge, so \[16 \times {10^{ - 3}}C\] charge flows across a cross-section per second.
Using the definition of electric current, we find total charge per second.
\[I = \dfrac{q}{t} \\ \]
\[\Rightarrow 16\,mA = \dfrac{q}{{1s}} \\ \]
\[\Rightarrow q = \left( {16 \times {{10}^{ - 3}}A} \right) \times \left( {1s} \right) \\ \]
\[\Rightarrow q = 16 \times {10^{ - 3}}C\]
So, the total charge striking the screen of the cathode ray tube is \[16 \times {10^{ - 3}}C\].
Let there be n electrons strike the screen of the cathode ray tube each second. Then the total charge striking the cathode ray tube per second will be,
\[q = ne\]
\[\Rightarrow q = 1.6 \times {10^{ - 19}}n\]
From the obtained charge,
\[1.6 \times {10^{ - 19}}n = 16 \times {10^{ - 3}} \\ \]
\[\Rightarrow n = \dfrac{{16 \times {{10}^{ - 3}}}}{{1.6 \times {{10}^{ - 19}}}} \\ \]
\[\Rightarrow n = {10^{17}}\]
Hence, \[{10^{17}}\] electrons strike the screen of the cathode ray per second when the current is 16 mA.
Therefore, the correct option is D.
Note: The electric current is a scalar quantity, so to find the rate of flow of charge we take the magnitude of the charge rather than considering whether charge is negative or positive. The charge on an electron is \[ - 1.6 \times {10^{ - 19}}C\] and the magnitude of the charge is \[1.6 \times {10^{ - 19}}C\].
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