What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.5*10^{15}Hz?
To calculate the kinetic energy of the emitted electrons when cesium is exposed to UV rays of a certain frequency, we need to use the equation for the energy of a photon and the equation for the kinetic energy of an electron.
The energy of a photon can be determined using the equation:
E = h * f
where E is the energy, h is the Planck's constant (approximately 6.626 x 10^-34 J.s), and f is the frequency of the UV rays.
Substituting the given frequency into the equation, we have:
E = (6.626 x 10^-34 J.s) * (1.5 x 10^15 Hz)
Next, we can equate the energy of the photon to the maximum kinetic energy of the ejected electron using the equation:
E = K.E. max
where K.E. max is the kinetic energy of the electron.
The equation for the kinetic energy of an electron is:
K.E. = (1/2) * m * v^2
where m is the mass of the electron (approximately 9.109 x 10^-31 kg), and v is the velocity of the electron.
Assuming that the ejected electron starts from rest, we can simplify the equation:
K.E. max = (1/2) * m * v^2
Now, we can rearrange the equation to solve for v:
v = sqrt((2 * K.E. max) / m)
By equating the energy of the photon (E) to the maximum kinetic energy of the ejected electron (K.E. max), we can solve for the velocity (v) of the electron. Once we know the velocity, we can find the kinetic energy using the equation for kinetic energy.
Note that we're assuming a simplified scenario here, neglecting factors like the work function and Fermi level of the metal. In reality, these factors influence the energy required to eject an electron from a metal surface.
Find the energy of the photon, subtract the work function of Cesium, then convert the remainder to KE.
http://hyperphysics.phy-astr.gsu.edu/HBASE/Tables/photoelec.html