Chapter 25: Q. 13 (page 709)

What potential difference is needed to accelerate an electron from rest to a speed of $2.0 \times 10^{6} m / s$ ?

Short Answer

Expert verified

The potential difference needed is $V_{1 \to 2} = 11.4 V$

Step by step solution

Given information and theory used

Given : To accelerate an electron from rest to a speed of : $2.0 \times 10^{6} m / s$

Theory used :

According to the law of conservation of energy, $U_{1} = U_{2} + \frac{1}{2} m v^{2}$ . ...(1)

Calculating potential difference needed

Allow the electron to accelerate from point 1 to point 2, where it will reach $v = 2.0 \times 10^{6} m / s$ .

Rewriting the law of conservation of energy, in equation (1) :

$\frac{1}{2} m v^{2} = - (U_{2} - U_{1})$

Through $(U_{2} - U_{1}) = q (V_{2} - V_{1}) = q V_{1 \to 2}$ , the potential difference is connected to the potential energy difference.

Because electron $q = - e$ , produces

$\frac{1}{2} m v^{2} = e V_{1 \to 2} \Rightarrow V_{1 \to 2} = \frac{m v^{2}}{2 e}$

We get $V_{1 \to 2} = 11.4 V$ by substituting $e = 1.60 \times 10^{- 19} C and m = 9.11 \times 10^{- 31} k g$

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What potential difference is needed to accelerate an electron from rest to a speed of $2.0 \times 10^{6} m / s$ ?

Short Answer

Step by step solution

Given information and theory used

Calculating potential difference needed

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What potential difference is needed to accelerate an electron from rest to a speed of $2.0 \times 10^{6} m / s$ ?