Question

An electron is accelerated through a potential difference of 1000 volts. Its velocity is nearly

• A. 3.8$\times 10^7$ m/s
• B. 1.9$\times 10^6$ m/s
• C. 1.9$\times 10^7$ m/s
• D. 5.7$\times 10^7$ m/s

Answer 1

Answer: (C)

1.9$\times 10^7$ m/s

Answer 2

The kinetic energy gained by the electron is equal to the work done on it by the electric field.

$\text{Kinetic Energy} = \text{Work Done}$

$\frac{1}{2}m_e v^2 = q_e V$

where $m_e$ is the mass of the electron, $v$ is its velocity, $q_e$ is the charge of the electron, and $V$ is the potential difference.

We are given $V = 1000$ volts.

The mass of an electron is $m_e \approx 9.11 \times 10^{-31}$ kg.

The charge of an electron is $q_e \approx 1.60 \times 10^{-19}$ C.

We need to find the velocity $v$.

From the equation, we can solve for $v$:

$v^2 = \frac{2 q_e V}{m_e}$

$v = \sqrt{\frac{2 q_e V}{m_e}}$

Substitute the values:

$v = \sqrt{\frac{2 \times (1.60 \times 10^{-19} \text{ C}) \times (1000 \text{ V})}{9.11 \times 10^{-31} \text{ kg}}}$

$v = \sqrt{\frac{3.20 \times 10^{-16}}{9.11 \times 10^{-31}}}$

$v = \sqrt{\frac{3.20}{9.11} \times 10^{15}}$

$v = \sqrt{0.35126 \times 10^{15}}$

To take the square root, we rewrite the exponent as an even number:

$v = \sqrt{3.5126 \times 10^{14}}$

$v = \sqrt{3.5126} \times 10^7$

Calculating the square root:

$\sqrt{3.5126} \approx 1.874$

So, $v \approx 1.874 \times 10^7$ m/s.

This value is nearly $1.9 \times 10^7$ m/s.