an electron in a photoelectron spectrometer is accelerated from rest by a uniform electric field to a speed of 420 km/s in 10 us. Determine the magnitude of the acceleration of electron in m/s2?



Answer :

Answer:

42×10⁹ m/s²

8.04×10⁻²⁰ Nm

8.04×10⁻²⁰ J

Explanation:

Acceleration (a) is the change in velocity (Δv) over time (Δt). Convert kilometers to meters, and microseconds to seconds, to calculate acceleration in meters per seconds squared.

[tex]\LARGE \text {$ a\ $}\huge \text {$ =\frac{\Delta v}{\Delta t} $}\\\\\LARGE \text {$ a\ $}\huge \text {$ =\frac{420\ km/s}{10\ \mu s} $}\\\\\LARGE \text {$ a\ $}\huge \text {$ =\frac{420\times10^3\ m/s}{10\times10^{-6}\ s} $}\\\\\LARGE \text {$ a=42\times10^9\ m/s^2 $}[/tex]

The kinetic energy (KE) is half the mass (m) times the square of the velocity (v). An electron has a mass of 9.11×10⁻³¹ kg.

[tex]\large \text {$ KE= $}\LARGE \text {$ \frac{1}{2} $}\large \text {$ mv^2 $}\\\\\large \text {$ KE= $}\LARGE \text {$ \frac{1}{2} $}\large \text {$ (9.11\times10^{-31}\ kg)\ (420\times10^3\ m/s)^2 $}\\\\\large \text {$ KE=803,502\times10^{-25}\ Nm $}\\\\\large \text {$ KE=8.04\times10^{-20}\ Nm $}[/tex]

According to the work-energy theorem, the work done on an object (W) is equal to the change in kinetic energy (ΔKE). Notice that a Newton-meter is the same as a Joule.

[tex]\large \text {$ W=\Delta KE $}\\\\\large \text {$ W=8.04\times10^{-20}\ J-0\ J $}\\\\\large \text {$ W=8.04\times10^{-20}\ J $}[/tex]