Question

What is the kinetic energy K of an electron with momentum 1.05×10−24 kilogram meters per second?Express your answer in electron volts to two significant figures.

Answer 1

3.7eV

Step-by-step explanation:

The kinetic energy (K) of a particle is the half the product of the mass (m) and the square of the velocity (v) of the particle. i.e

K =
`(1)/(2)` x m x v² --------------------(i)

Also, the momentum (p) of a particle is the product of the mass and the velocity of the particle. i.e

p = m x v [square both sides]

p² = (mv)²

p² = m²v² [make v² subject of the formula]

v² =
`(p^(2) )/(m^(2) )`

Now substitute v² =
`(p^(2) )/(m^(2) )` into equation (i) as follows;

K =
`(1)/(2)` x m x
`(p^(2) )/(m^(2) )`

K =
`(1)/(2)` x
`(p^(2) )/(m)` --------------------------(ii)

From the question the particle is an electron with the following details;

m = mass of the electron = 9.1 x 10⁻³¹kg [known value]

p = momentum of the electron = 1.05 x 10⁻²⁴kgm/s [given value]

Substitute these values into equation (ii) as follows;

K =
`(1)/(2)` x
`((1.05 * 10^(-24) )^(2) )/(9.1 * 10^(-31))`

K =
`(1)/(2)` x
`((1.10 * 10^(-48) ) )/(9.1 * 10^(-31))`

K =
`(1)/(2)` x 0.12 x 10⁻¹⁷

K = 0.06 x 10⁻¹⁷

K = 6.00 x 10⁻¹⁹ J

Convert the result to electron volts (eV) as follows;

1 J = 6.2 x 10¹⁸ eV

6.00 x 10⁻¹⁹ J = 6.00 x 10⁻¹⁹ J x 6.2 x 10¹⁸ eV / 1 J = 3.7eV

Therefore, the kinetic energy K of an electron with that momentum is 3.7eV

Answer 2

Momentum = mvwhere m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m = (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2 = (9.1×10∧-31 × 1,153,846.154∧2) ÷2 = (1.21154×10∧-18) ÷ 2 = 6.05769×10∧-19 J

Thus, 3.8eV since 1J=6. 242×10^18