Recent content by NATSALANE

an electron travels at 1 x 106 m/s. when it hits a target, its speed decrease in half and a photo is released. find the phton's wavelength. so far i know E photon = (hc) / lamda what about the speed? does the photons' energy same as kinetic energy of the electron which is just 1/2 mv ^2...

okay then. so EK = (hc)/λ - eVo i got the same answer as W but opposite sign (negative)...

all i am given is the wavelength and cutoff potential, so in order to find the maximum kin energy, don't i use this equation: Kinetic energy = speed of electron * cutoff potential ? how does that equate to work?

also, how do i graph the maximum kinetic energy-vs-frqeuency of photons graph? i only have one value of frequency and kinetic energy. i know that the slope must be the same as the value of plank's constant, but assuming i do not know the constant, how can i find more points on the line to connect?

given the cutoff potential (0.25 V) and wavelength (578 nm), how do i find the maximum kinetic energy of the electrons ejected from photoelectrif surface, in both eV and J? my guess is, KE = e Vo = (1.6 x 10-19) x (0.25 V) = 4.0 x 10-20 J but if i convert that to eV, i get the same...

The direction is horizontal to conductor. If that's right, maybe you can help me with another question: Two sources that are placed 2.0 m apart operate at a frequency of 1.0 Hz. If the waves are 0.60 m, at what angles (from the centre line of the interference pattern) are nodal lines...

Using the right-hand rule, the thumb should be pointing up. Not sure which way to point the fingers, though... Would the direction of magnetic force be westward?

This is another part where I am stuck. Since the magnetic force is perpendicular to both the velocity and the magnetic field B, would the direction of the force be towards the observer?

Thank you for your help. This is what I arrived at: I (current) = 18 A [upward] v (speed) = 8.92 x 10^4 m/s r (radius) = 0.2 m q (charge) = 1.602 x 10^-19 C B = μ0 [I/(2πr)) = 4π x 10^-7 [18 / (2 x π x 0.20) = 1.8 x 10^-5 F nag. = qvB sinΘ = (1.602x10^-19) x (8.92 x 10^4) x...

Homework Statement In a vacuum, a straight conductor has 18-A current that goes upward. An electron is traveling at 8.92 x 10^4 m/s. If the electron is 0.2 m from the conductor and its instantaneous velocity is parallel to the conductor (albeit heading downward), find the magnitude and...