Answer:
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
3.7eV
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 = [tex]\frac{1}{2}[/tex] 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 = [tex]\frac{p^{2} }{m^{2} }[/tex]
Now substitute v² = [tex]\frac{p^{2} }{m^{2} }[/tex] into equation (i) as follows;
K = [tex]\frac{1}{2}[/tex] x m x [tex]\frac{p^{2} }{m^{2} }[/tex]
K = [tex]\frac{1}{2}[/tex] x [tex]\frac{p^{2} }{m}[/tex] --------------------------(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 = [tex]\frac{1}{2}[/tex] x [tex]\frac{(1.05 * 10^{-24} )^{2} }{9.1 * 10^{-31}}[/tex]
K = [tex]\frac{1}{2}[/tex] x [tex]\frac{(1.10 * 10^{-48} ) }{9.1 * 10^{-31}}[/tex]
K = [tex]\frac{1}{2}[/tex] 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
