Answer:
![W = 462.5 keV](https://tex.z-dn.net/?f=W%20%3D%20462.5%20keV)
Explanation:
As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron
So here we have to find the work done by electric field on moving electron
So we have
![F = qE](https://tex.z-dn.net/?f=F%20%3D%20qE)
![F = (1.6 \times 10^{-19})(1.85 \times 10^6)](https://tex.z-dn.net/?f=F%20%3D%20%281.6%20%5Ctimes%2010%5E%7B-19%7D%29%281.85%20%5Ctimes%2010%5E6%29)
![F = 2.96 \times 10^{-13} N](https://tex.z-dn.net/?f=F%20%3D%202.96%20%5Ctimes%2010%5E%7B-13%7D%20N)
now the distance moved by the electron is given as
![d = 0.25 m](https://tex.z-dn.net/?f=d%20%3D%200.25%20m)
so we have
![W = F.d](https://tex.z-dn.net/?f=W%20%3D%20F.d)
![W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)](https://tex.z-dn.net/?f=W%20%3D%20%281.6%20%5Ctimes%2010%5E%7B-19%7D%29%281.85%20%5Ctimes%2010%5E6%29%280.25%29)
![W = 7.4 \times 10^{-14} J](https://tex.z-dn.net/?f=W%20%3D%207.4%20%5Ctimes%2010%5E%7B-14%7D%20J)
now we have to convert it into keV units
so we have
![1 keV = 1.6 \times 10^{-16} J](https://tex.z-dn.net/?f=1%20keV%20%3D%201.6%20%5Ctimes%2010%5E%7B-16%7D%20J)
![W = 462.5 keV](https://tex.z-dn.net/?f=W%20%3D%20462.5%20keV)
I hope this helps you out!
I don’t fully understand this question but imma say more sun in summer
The light collection of a telescope is proportional to the AREA of the objective lens or mirror. So the 8m collects 4 times as much light as the 4m does in the same time. Or the bigger one collects the same amount of light in 1/4 of the time that it takes the smaller one.
Under the same observing conditions that the 4m needs 3 hours, the 8m monster can do the job in 45 minutes.